Is there an exact or good approximate expression for the expectation, variance or other moments of the maximum of $n$ independent, identically distributed gaussian random variables where $n$ is large?

If $F$ is the cumulative distribution function for a standard gaussian and $f$ is the probability density function, then the CDF for the maximum is (from the study of order statistics) given by

$$F_{\rm max}(x) = F(x)^n$$

and the PDF is

$$f_{\rm max}(x) = n F(x)^{n-1} f(x)$$

so it's certainly possible to write down integrals which evaluate to the expectation and other moments, but it's not pretty. My intuition tells me that the expectation of the maximum would be proportional to $\log n$, although I don't see how to go about proving this.

The $\max$-central limit theorem (Fisher-Tippet-Gnedenko theorem) can be used to provide a decent approximation when $n$ is large. See this example at reference page for extreme value distribution in Mathematica.
The $\max$-central limit theorem states that $F_\max(x) = \left(\Phi(x)\right)^n \approx F_{\text{EV}}\left(\frac{x-\mu_n}{\sigma_n}\right)$, where $F_{EV} = \exp(-\exp(-x))$ is the cumulative distribution function for the extreme value distribution, and
$$
   \mu_n = \Phi^{-1}\left(1-\frac{1}{n} \right) \qquad \qquad
   \sigma_n = \Phi^{-1}\left(1-\frac{1}{n} \cdot \mathrm{e}^{-1}\right)- \Phi^{-1}\left(1-\frac{1}{n} \right)
$$
Here $\Phi^{-1}(q)$ denotes the inverse cdf of the standard normal distribution.
The mean of the maximum of the size $n$ normal sample, for large $n$, is well approximated by
$$ \begin{eqnarray}
  m_n &=&  \sqrt{2} \left((\gamma -1) \Phi^{-1}\left(2-\frac{2}{n}\right)-\gamma  \Phi^{-1}\left(2-\frac{2}{e n}\right)\right) \\ &=& \sqrt{\log \left(\frac{n^2}{2 \pi  \log \left(\frac{n^2}{2\pi} \right)}\right)} \cdot \left(1 + \frac{\gamma}{\log (n)} + \mathcal{o} \left(\frac{1}{\log (n)} \right) \right) 
\end{eqnarray}$$
where $\gamma$ is the Euler-Mascheroni constant.
