The question reads to me like the OP was asking when $U = (X,Y,Z)^{\mathrm{T}}$ are jointly normal then what is the probability $P(X \geq Y \mbox{ and } X \geq Z)$?
For that question we could look at the joint distribution of $AU$ where $A$ looks like
$$
A=\left[
\begin{array}{ccc}
1 & -1 & 0 \newline
1 & 0 & -1
\end{array}\right]
$$
Of course, $AU$ is also jointly normal with mean $A\mu$ and variance-covariance $A\Sigma A^{\mathrm{T}}$, and the desired probability is $P(AU > \mathbf{0}_{n-1})$. We could get this in R with something like
set.seed(1)
Mu <- c(1,2,3)
library(MCMCpack)
S <- rwish(3, diag(3)) # get var-cov matrix
A <- matrix(c(1,-1,0, 1,0,-1), nrow = 2, byrow = TRUE)
newMu <- as.vector(A %*% Mu)
newS <- A %*% S %*% t(A)
library(mvtnorm)
pmvnorm(lower=c(0,0), mean = newMu, sigma = newS)
which is about 0.1446487 on my system. If a person knew something about the matrix $\Sigma$ then (s)he might even be able to write something down that looks like a formula (I haven't tried, though).
Best Answer
There is a section entitled 'Bivariate Half-normal distribution in: Continuous Multivariate Distributions: Models and applications By Samuel Kotz, Norman Lloyd Johnson, N. Balakrishnan.
I would be curious to see how this can be generalized to a random vector of any dimensions.
In fact, the bivariate case appears to be thoroughly treated in this paper: http://www.stat-athens.aueb.gr/~jpan/papers/Panaretos-ApplStatScience2001(119-136)ft.pdf