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, even the $n=2$ integral seems intractable. 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 $$E(X_{\rm max})=\mu+\sigma/\sqrt\pi,\;\;{\rm Var}\,(X_{\rm max})=(1-1/\pi)\sigma^2.$$
If you are satisfied with a large-$n$ approximation, see this MSE posting.