The classical statement of the Central Limit Theorem (CLT) considers a sequence of independent, identically distributed random variables $X_1, X_2, \ldots, X_n, \ldots$ with common distribution $F$. This sequence models the situation we confront when designing a sampling program or experiment: if we can obtain $n$ independent observations of the same underlying phenomenon, then the finite collection $X_1, X_2, \ldots, X_n$ models the anticipated data. Allowing the sequence to be infinite is a convenient way to contemplate arbitrarily large sample sizes.
Various laws of large numbers assert that the mean
$$m(X_1, X_2, \ldots, X_n) = \frac{1}{n}(X_1 + X_2 + \cdots + X_n)$$
will closely approach the expectation of $F$, $\mu(F)$, with high probability, provided $F$ actually has an expectation. (Not all distributions do.) This implies the deviation $m(X_1, X_2, \ldots, X_n) - \mu(F)$ (which, as a function of these $n$ random variables, is also a random variable) will tend to get smaller as $n$ increases. The CLT adds to this in a much more specific way: it states (under some conditions, which I will discuss below) that if we rescale this deviation by $\sqrt{n}$, it will have a distribution function $F_n$ that approaches some zero-mean Normal distribution function as $n$ grows large. (My answer at https://stats.stackexchange.com/a/3904 attempts to explain why this is and why the factor of $\sqrt{n}$ is the right one to use.)
This is not a standard statement of the CLT. Let's connect it with the usual one. That limiting zero-mean Normal distribution will be completely determined by a second parameter, which is usually chosen to be a measure of its spread (naturally!), such as its variance or standard deviation. Let $\sigma^2$ be its variance. Surely it must have some relationship to a similar property of $F$. To discover what this might be, let $F$ have a variance $\tau^2$--which might be infinite, by the way. Regardless, because the $X_i$ are independent, we easily compute the variance of the means:
$$\eqalign{
\text{Var}(m(X_1, X_2, \ldots, X_n)) &= \text{Var}(\frac{1}{n}(X_1 + X_2 + \cdots + X_n)) \\
&= \left(\frac{1}{n}\right)^2(\text{Var}(X_1) + \text{Var}(X_2) + \cdots + \text{Var}(X_n)) \\
&= \left(\frac{1}{n}\right)^2(\tau^2 + \tau^2 + \cdots + \tau^2) \\
&= \frac{\tau^2}{n}.
}$$
Consequently, the variance of the standardized residuals equals $\tau^2/n \times (\sqrt{n})^2 = \tau^2$: it is constant. The variance of the limiting Normal distribution, then, must be $\tau^2$ itself. (This immediately shows that the theorem can hold only when $\tau^2$ is finite: that is the additional assumption I glossed over earlier.)
(If we had chosen any other measure of spread of $F$ we could still succeed in connecting it to $\sigma^2$, but we would not have found that the corresponding measure of spread of the standardized mean deviation is constant for all $n$, which is a beautiful--albeit inessential--simplification.)
If we had wished, we could have standardized the mean deviations all along by dividing them by $\tau$ as well as multiplying them by $\sqrt{n}$. That would have ensured the limiting distribution is standard Normal, with unit variance. Whether you elect to standardize by $\tau$ in this way or not is really a matter of taste: it's the same theorem and the same conclusion in the end. What mattered was the multiplication by $\sqrt{n}$.
Note that you could multiply the deviations by some factor other than $\sqrt{n}$. You could use $\sqrt{n} + \exp(-n)$, or $n^{1/2 + 1/n}$, or anything else that asymptotically behaves just like $\sqrt{n}$. Any other asymptotic form would, in the limit, reduce $\sigma^2$ to $0$ or blow it up to $\infty$. This observation refines our appreciation of the CLT by showing the extent to which it is flexible concerning how the standardization is performed. We might want to state the CLT, then, in the following way.
Provided the deviation between the mean of a sequence of IID variables (with common distribution $F$) and the underlying expectation is scaled asymptotically by $\sqrt{n}$, this scaled deviation will have a zero-mean Normal limiting distribution whose variance is that of $F$.
Although variances are involved in the statement, they appear only because they are needed to characterize the limiting Normal distribution and relate its spread to that of $F$. This is only an incidental aspect. It has nothing to do with variance being "best" in any sense. The crux of the matter is the asymptotic rescaling by $\sqrt{n}$.