For independently distributed $x_i$'s, each with cumulative distribution $$F_i(x_i)=\tfrac{1}{2}+\tfrac{1}{2}{\rm Erf}\,[(x_i-\mu_i)/(\sigma_i\sqrt 2],$$ 
the cumulative distribution of the maximum is given by
$$P({\rm max}_i \,x_i<X_{\rm max})=\prod_{i=1}^n P(x_i<X_{\rm max})=\prod_{i=1}^n F_i(X_{\rm max}).$$
For small $n$ you can now calculate moments of $X_{\rm max}$ by integration,
$$E(X_{\rm max}^p)=\int_{-\infty}^\infty x^p\frac{d}{dx}\left(\prod_{i=1}^n F_i(x)\right)\,dx.$$
There is unlikely to be a closed-form answer for arbitrary $n$, in fact, even the $n=2$ integral seems problematic (*Mathematica* fails to evaluate it). If you take the $\mu_i$'s and $\sigma_i$'s to be the same, then progress can be made, for $n=2$ I find
$$E(X_{\rm max})=\mu+\sigma/\sqrt\pi,\;\;{\rm Var}\,(X_{\rm max})=(1-1/\pi)\sigma^2.$$

Perhaps you are satisfied with a *large-$n$ approximation*. For identical $\mu_i$'s and $\sigma_i$'s it is given by the <A HREF="https://en.wikipedia.org/wiki/Fisher–Tippett–Gnedenko_theorem">Fisher–Tippett–Gnedenko theorem</A>, see for example  <A HREF="https://math.stackexchange.com/questions/89030/expectation-of-the-maximum-of-gaussian-random-variables">this MSE posting.</A> I have found one paper that generalizes this to arbitrary $\mu_i$'s and $\sigma_i$'s: <A HREF="http://www.sdssu.edu.ph/files/Research/14.THE%20DISTRIBUTION%20OF%20THE%20MAXIMUM%20OF%20INDEPENDENT%20NORMAL%20RANDOM%20VARIABLES%20IID%20AND%20INID%20CASES.pdf">On the distribution of the maximum of n independent normal random variables: iid and inid cases,</A> but I have difficulty parsing their result.

<sub>There is more in that referene that I do not understand. They give the inequality
$$\frac{1}{n}\sum_i\mu_i\leq E(X_{\rm max})\leq \frac{1}{n}\sum_i\mu_i+\frac{n-1}{n}\,{\rm max}_i\,\mu_i$$
which contradicts the $n=2$ result given above.</sub>