[My answer to the second question you list](http://stats.stackexchange.com/a/43705/6633) 
has a simple form of the more general result given by @whuber, but is readily adapted
to the general case.  Instead of
$$P(X_1 > \max X_i  \mid X_1 = \alpha) = \prod_{i=2}^n P\{X_i < \alpha \mid X_1 = \alpha\} 
= \left[\Phi(\alpha)\right]^{n-1}$$
which applies when the $X_i$ are independent $N(0,1)$ random variables, we have
$$P(X_1 > \max X_i  \mid X_1 = \alpha) = \prod_{i=2}^n P\{X_i < \alpha \mid X_1 = \alpha\} 
= \prod_{i=2}^n \Phi\left(\frac{\alpha-\mu_i}{\sigma_i}\right)$$
since the $X_i$ are independent $N(\mu_i, \sigma_i^2)$ random variables, and instead of 
$$P(X_1 > \max X_i)  
= \int_{-\infty}^{\infty}\left[\Phi(\alpha)\right]^{n-1}
\phi(\alpha-\mu)\,\mathrm d\alpha$$
we have
$$P(X_1 > \max X_i)  
= \int_{-\infty}^{\infty}\prod_{i=2}^n \Phi\left(\frac{\alpha-\mu_i}{\sigma_i}\right)
\frac{1}{\sigma}\phi\left(\frac{\alpha-\mu_1}{\sigma_1}\right)\,\mathrm d\alpha$$
where $\Phi(\cdot)$ and $\phi(\cdot)$ are the cumulative distribution function
and probability density function of the standard normal random variable.
This is just whuber's answer expressed in different notation.  

The _complementary_ probability 
$P(X_1 < \max X_i) = P\{(X_1 < X_2) \cup \cdots \cup (X_1 < X_n)$ can also
be bounded above by the union bound discussed in my answer to the other question.
We have that
$$\begin{align*}
P(X_1 < \max X_i) &=  P\{(X_1 < X_2) \cup \cdots \cup (X_1 < X_n)\\
&\leq \sum_{i=2}^n P(X_1 < X_i)\\
&= \sum_{i=2}^n Q\left(\frac{\mu_1 - \mu_i}{\sqrt{\sigma_1^2 + \sigma_i^2}}\right)
\end{align*}$$
since $X_i-X_1 \sim N(\mu_i-\mu_1,\sigma_i^2+\sigma_1^2)$.  Note that
$Q(x) = 1-\Phi(x)$ is the _complementary_  standard normal
distribution function. The union bound is very tight when
 $\mu_1 \gg \max \mu_i$ and the variances are roughly
comparable even for large $n$, but for small $n$, the bound
can exceed $1$ and thus be useless.