Given a Gaussian random vector of length N:
$$ X \sim N(\mu,\Sigma) $$
where $\Sigma$ is non-diagonal, what is the probability that a chosen component of $X$ is larger than any other? i.e.
$$P(X_n > max(X_i)) \; \text{ for } \; i = [1,2, ..., n-1,n+1,...,N]$$ For a diagonal covariance matrix, i.e. vector with independent components, we can solve the problem as shown here. I'm not sure how to approach this for the general case, though.
Furthermore -- how can the answer to be extended to include a group of components? e.g.:
$$P(max(X_{n}, X_{n+1}, X_{n+2}) > max(X_i)) \; \text{ for } \; i = [1,2, ..., n-1,n+3,...,N]$$
Expanding on @Max's comment: $Y=X_n-X$ is itself a Gaussian random vector whose covariance matrix can easily be worked out from that of $X$. Then we are looking for the probability that all components of $Y$ are positive, which is an "orthant probability"; closed formulas (for small $N$ only), recursion formulas, and approximation schemes can be found in the literature. See for example: R. H. Bacon, Approximations to multivariate normal orthant probabilities, The Annals of Mathematical Statistics 34 (1963), no. 1, 191–198.