W.I.P.: Work in progress
Following p. 370 of Cramer's 1946 Mathematical Methods of Statistics, define $$\Xi_n = n(1 - \Phi(Z_n)) \,. $$ Here $\Phi$ is the cumulative distribution function of the standard normal distribution, $\mathscr{N}(0,1)$. As a consequence of its definition, we are guaranteed that $0\le \Xi_n \le n$ almost surely.
Consider a given realization $\omega \in \Omega$ of our sample space. Then in this sense $Z_n$ is both a function of $n$ and $\omega$, and $\Xi_n$ a function of $Z_n, n$, and $\omega$. For a fixed $\omega$, we can consider $Z_n$ a deterministic function of $n$, and $\Xi_n$ a deterministic function of $Z_n$ and $n$, thereby simplifying the problem. We aim to show results which hold for almost surely all $\omega \in \Omega$, allowing us to transfer our results from a non-deterministic analysis to the non-deterministic setting.
Following p. 374 of Cramer's 1946 Mathematical Methods of Statistics, assume for the moment (I aim to come back and supply a proof later) that we are able to show that (for any given $\omega \in \Omega$) the following asymptotic expansion holds (using integration by parts and the definition of $\Phi$):
$$ \frac{\sqrt{2\pi}}{n}\Xi_n = \frac{1}{Z_n}e^{-\frac{Z_n^2}{2}}\left( 1 + O \left( \frac{1}{Z_n^2} \right) \right) \quad ~~ as ~~ Z_n \to \infty \,. \tag{~}$$
Clearly we have that $Z_{n+1} \ge Z_n$ for any $n$, and $Z_n$ is almost surely an increasing function of $n$ as $n\to \infty$, therefore we claim in what follows throughout that for (almost surely all) fixed $\omega$: $$ Z_n \to \infty \quad \iff \quad n \to \infty \,. $$
Hence it follows that we have (where $\sim$ denotes asymptotic equivalence):
$$ \frac{\sqrt{2\pi}}{n} \Xi_n \sim \frac{1}{Z_n} e^{-\frac{1}{Z_n^2}} \quad ~~ as ~~ Z_n \to \infty \quad n \to \infty \,. $$
How we proceed in what follows amounts essentially to the method of dominant balance, and our manipulations will be formally justified by the following lemma:
Lemma: Assume that $f(n) \sim g(n)$ as $n \to \infty$, and $f(n) \to \infty$ (thus $g(n) \to \infty$). Then given any function $h$ which is formed via compositions, additions, and multiplications of logarithms and power laws (essentially any "polylog" function), we must have also that as $n \to \infty$: $$ h(f(n)) \sim h(g(n)) \,. $$ In other words, such "polylog" functions preserve asymptotic equivalence.
The truth of this lemma is a consequence of Theorem 2.1. as written here. Note also that what follows is mostly an expanded (more details) version of the answer to a similar question found here.
Taking logarithms of both sides, we get that:
$$\log ( \sqrt{2\pi} \Xi_n ) - \log n \sim -\log Z_n - \frac{Z_n^2}{2} \,. \tag{1}$$
This is where Cramer is somewhat cagey; he just says "assuming $\Xi_n$ is bounded", we can conclude blah blah blah. But showing that $\Xi_n$ is suitably bounded almost surely seems to be actually somewhat non-trivial. It seems that the proof of this may essentially be part of what's discussed on pp. 265-267 of Galambos, but I am not sure given that I am still working to understand the content of that book.
Anyway, assuming one can show that $\log \Xi_n = o(\log n)$, presumably via $\Xi_n = o(n)$ almost surely, then it follows (since the $-Z_n^2/2$ term dominates the $-\log Z_n$ term) that:
$$ - \log n \sim - \frac{Z_n^2}{2} \quad \implies \quad Z_n \sim \sqrt{2 \log n} \,. $$
This is somewhat nice, since it is already most of what we want to show, although again it is worthwhile to note that it is essentially only kicking the can down the road, since now we have to show some certain almost surely boundedness of $\Xi_n$. On the other hand, $\Xi_n$ has the same distribution for any maximum of i.i.d. continuous random variables, so this may be tractable.
Anyway, if $Z_n \sim \sqrt{2 \log n}$ a.s., then clearly one can also conclude that $Z_n \sim \sqrt{2 \log n}(1 + \alpha(n))$ for any $\alpha(n)$ which is $o(1)$ as $n \to \infty$. Using our lemma about polylog functions preserving asymptotic equivalence above, we can substitute this expression back into $(1)$ to get:
$$\log(\sqrt{2 \pi} \Xi _n)- \log n \sim -\log (1 + \alpha) - \frac{1}{2}\log 2 - \frac{1}{2}\log \log n - \log n - 2 \alpha \log n - \alpha^2 \log n \,. $$
$$ \implies -\log(\Xi_n \sqrt{2 \pi}) \sim \log(1 + \alpha) + \frac{1}{2} \log 2 + \frac{1}{2} \log \log n + 2\alpha \log n + \alpha^2 \log n \,. $$
Here we have to go even further, and assume that $\log \Xi_n = o( \log \log n) ~~ as ~~ n \to \infty$, presumably via $\Xi_n = o(\log n)$, almost surely. Again, all Cramer says is "assuming $\Xi_n$ is bounded". But since all one can say a priori about $\Xi_n$ is that $0 \le Xi_n \le n$ a.s., it hardly seems clear that one should have $\Xi_n = O(1)$ almost surely, which seems to be the substance of Cramer's claim.
But anyway, assuming one believes that, then it follows that the dominant term which does not contain $\alpha$ is $\frac{1}{2} \log \log n$. Since $\alpha = o(1)$, it follows that $\alpha^2 = o(\alpha)$, and clearly $\log ( 1 + \alpha) = o (\alpha) = o(o(\alpha \log n))$, so the dominant term containing $\alpha$ is $2 \alpha \log n$. Therefore, we can rearrange and (dividing everything by $\frac{1}{2}\log\log n$ or $2 \alpha \log n$) find that
$$ - \frac{1}{2} \log \log n \sim 2 \alpha \log n \quad \implies \quad \alpha \sim - \frac{\log \log n}{4 \log n} \,. $$
Therefore, substituting this back into the above, we get that:
$$Z_n \sim \sqrt{2 \log n}- \frac{\log\log n}{2 \sqrt{2 \log n}} \,, $$
again, assuming we believe certain things about $\Xi_n$.
We rehash the same technique again; since $Z_n \sim \sqrt{2 \log n} - \frac{\log \log n}{2 \sqrt{2 \log n}}$, then it also follows that
$$ Z_n \sim \sqrt{2 \log n} - \frac{\log \log n}{2 \sqrt{2 \log n}} (1 + \beta(n)) = \sqrt{2 \log n} \left( 1 - \frac{\log \log n}{8 \log n}(1 + \beta(n)) \right) \,,$$
when $\beta(n)=o(1)$. Let's simplify a little before substituting directly back into (1); we get that:
$$ \log Z_n \sim \log(\sqrt{2 \log n}) + \underbrace{\log \left(1 - \frac{\log \log n}{8 \log n}(1 + \beta(n)) \right) }_{\log(O(1)) = o(\log n)} \sim \log (\sqrt{2 \log n}) \,.$$
$$ \frac{Z_n^2}{2} \sim \log n - \frac{1}{2} \log \log n (1 + \beta) + \underbrace{\frac{(\log \log n)^2}{8 \log n} ( 1 \beta)^2}_{o((1+ \beta) \log \log n)} \sim \log n - \frac{1}{2} (1 + \beta) \log \log n \,. $$
Substituting this back into (1), we find that:
$$ \log ( \sqrt{2 \pi} \Xi_n) - \log n \sim - \log(\sqrt{2 \log n}) - \log n + \frac{1}{2}(1 + \beta) \log \log n \quad \implies \quad \beta \sim \frac{\log (4 \pi \Xi_n^2)}{\log \log n} \,. $$
Therefore, we conclude that almost surely
$$Z_n \sim \sqrt{2 \log n} - \frac{\log \log n}{2 \sqrt{2 \log n}} \left(1 + \frac{\log(4 \pi) + 2 \log( \Xi_n)}{\log \log n} \right)\\ = \sqrt{2 \log n} - \frac{\log \log n + \log (4 \pi)}{ 2 \sqrt{2 \log n} } - \frac{\log (\Xi_n)}{\sqrt{2 \log n}} \,. $$
This corresponds to the final result on p.374 of Cramer's 1946 Mathematical Methods of Statistics except that here the exact order of the error term isn't given. Apparently applying this one more term gives the exact order of the error term, but anyway it doesn't seem necessary to prove the results about the maxima of i.i.d. standard normals in which we are interested.
Given the result of the above, namely that almost surely:
$$Z_n \sim \sqrt{2 \log n} - \frac{\log \log n + \log (4 \pi)}{2 \sqrt{2 \log n}} - \frac{\log (\Xi_n)}{\sqrt{2 \log n}} \quad \implies \\ Z_n = \sqrt{2 \log n} - \frac{\log \log n + \log (4 \pi)}{2 \sqrt{2 \log n}} - \frac{\log (\Xi_n)}{\sqrt{2 \log n}} + o(1)\,. \tag{$\dagger$}$$
2. Then by linearity of expectation it follows that:
$$ \mathbb{E}Z_n = \sqrt{2 \log n} - \frac{\log \log n + \log (4 \pi)}{2 \sqrt{2 \log n}} - \frac{\mathbb{E}[\log (\Xi_n)]}{\sqrt{2 \log n}} + o(1) \quad \implies \\ \frac{\mathbb{E}Z_n}{\sqrt{2 \log n}} = 1 - \frac{\mathbb{E}[\log \Xi_n]}{2 \log n} + o(1) \,. $$
Therefore, we have shown that
$$ \lim_{n \to \infty } \frac{\mathbb{E} Z_n}{\sqrt{2 \log n}} = 1 \,,$$
as long as we can also show that
$$ \mathbb{E}[\log \Xi_n] = o(\log n) \,. $$
This might not be too difficult to show since again $\Xi_n$ has the same distribution for every continuous random variable. Thus we have the second result from above.
1. Similarly, we also have from the above that almost surely:
$$\frac{Z_n}{\sqrt{2 \log n}} = 1 - \frac{\log(\Xi_n)}{2 \log n} +o(1),.$$
Therefore, if we can show that:
$$ \log(\Xi_n) = o(\log n) \text{ almost surely}, \tag{*}$$
then we will have shown the first result from above. Result (*) would also clearly imply a fortiori that $\mathbb{E}[\log (\Xi_n)] = o(\log n)$, thereby also giving us the first result from above.
Also note that in the proof above of ($\dagger$) we needed to assume anyway that $\Xi_n = o(\log n)$ almost surely (or at least something similar), so that if we are able to show ($\dagger$) then we will most likely also have in the process needed to show $\Xi_n = o(\log n)$ almost surely, and therefore if we can prove $(\dagger)$ we will most likely be able to immediately reach all of the following conclusions.
3. However, if we have this result, then I don't understand how one would also have that $\mathbb{E}Z_n = \sqrt{2 \log n} + \Theta(1)$, since $o(1) \not= \Theta(1)$. But at the very least it would seem to be true that $$\mathbb{E}Z_n = \sqrt{2 \log n} + O(1) \,.$$
So then it seems that we can focus on answering the question of how to show that $$ \Xi_n = o(\log n) \text{ almost surely.} $$
(Seemingly this would be equivalent to $\log(\Xi_n) = o(\log \log n)$ almost surely via the continuity of $\log/\exp$ and the continuous mapping theorem, but I have not thought about it enough to be sure.)
We will also need to do the grunt work of providing a proof for (~), but to the best of my knowledge that is just calculus and involves no probability theory, although I have yet to sit down and try it yet.