For independently distributed $x_i$'s, each with cumulative distribution $F_i(x_i)$, 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, I have note even succeeded in evaluating the $n=2$ integral. 

If you are satisfied with a large-$n$ approximation, see <A HREF="https://math.stackexchange.com/questions/473229/expected-value-for-maximum-of-n-normal-random-variable">this MSE posting.</A>