What is the probability that random variable $x_1$ is maximum of random vector $X=(x_i)$ from a multivariate normal distribution? Given a $n$-dimensional multivariate normal distribution $X=(x_i) \sim \mathcal{N}(\mu, \Sigma)$ with mean $\mu$ and covariance matrix $\Sigma$, what is the probability that $\forall j\in {1,\ldots,n}:x_1 \geq x_j$?
 A: Answer updated thanks to remarks from Whuber and Srikant

Proposition
  Let C=[C_1;C_2] be a 2*n matrix, $X^0=(X^0_i)\sim \mathcal{N} (0,\Sigma)$ $\mathbb{R}^n$ valued. Let $\Sigma^Y=^tC\Sigma C=(\sigma^Y_{ij})$. Then, for $u_1,u_2\in\mathbb{R}$
$P(^tC_1X^0\geq u_1\text{ and } ^tC_2X^0\geq u_2)=\mathbb{E}\left [ \bar{\Phi}\left (\frac{u_2-\frac{\sigma^Y_{21}}{\sigma^Y_{11}}^tC_1X^0}{\sqrt{\sigma^Y_{22}-\frac{\sigma^Y_{21}\sigma^Y_{12}}{\sigma^Y_{11}}}} \right )1_{ ^tC_1X^0\geq u_1 } \right ]$ 
where $\bar{\Phi}=P(\mathcal{N}(0,1)>z)$

Answer to the question when the dimension is 3
Assume $i=1$, $\Sigma=(\sigma_{ij})$. 
The probability $P(X_1>X_2 \text{ and }X_1>X_3)$ is obtained using the preceding proposition with $X^0=X-\mu$, $C_1=(1,-1,0)$, $C_2=(1,0,-1)$, $u_1=\mu_2-\mu1$ and $u_2=\mu_3-\mu1$. This gives
$\sigma^Y_{11}=\sigma_{11}+\sigma_{22}-2\sigma_{12}$
$\sigma^Y_{22}=\sigma_{11}+\sigma_{33}-2\sigma_{13}$
$\sigma^Y_{12}=\sigma_{11}+2\sigma_{23}-\sigma_{31}-\sigma_{21}$
Proof of the proposition
Assume $c\in\mathbb{R}^n$ and $\Sigma$ has full rank. It is easy to show that for any $u\in\mathbb{R}$
$$P(^tcX^0>u)=\bar{\Phi} \left (\frac{u}{\|\Sigma^{1/2}c\|_2} \right )$$
Let us denote  $Y_1=^tC_1X^0,Y_2=^tC_2X^0$. From the correlation theorem, since $Y=(Y_1,Y_2)$ is centered gaussian in $\mathbb{R}^2$  with covariance $\Sigma^Y$
then $Y_2|Y_1$ is gaussian with mean $\frac{\sigma^Y_{21}}{\sigma^Y_{11}}Y_1$ and variance  $\sigma^Y_{22}-\frac{\sigma^Y_{21}}{\sigma^Y_{11}}\sigma^Y_{12}$. 
This, with
$P(Y_1>u_1 \text{ and }  Y_2>u_2)=\mathbb{E}\left [\mathbb{E}[1_{Y_2\geq u_2 }|Y_1]  1_{Y_1\geq u_1 }\right ] $
gives the desired result. 
How to extend the proposition
If we want to be able to solve the initial problem with dimension larger than $3$, 
we need to compute
$P(\forall j \; ^tc_jX^0\geq u_j) $
(for well chosen $u_j$). Set $Y=(Y_1,\dots,Y_n)$ with $Y_j=^tc_jX$ centered $\mathbb{R}$-valued gaussians.
You can use the correlation theorem iteratively to derive the distribution of $Y_1|Y_{2:n}$, $Y_2|Y_{3:n}$..... This may give something like a  recurcive formulation of the solution to the proposition when C is $p*n$ (recurcive on $p$). 
A: I interpreted the question to ask the distribution of the maximal element of a multivariate normal. In this case, the CDF can be computed from the CDF of a multivariate normal. This usually doesn't have a nice solution (even in terms of the univariate normal CDF), however can be evaluated numerically. In R:
library(mvtnorm)
# given xl, mu and sigma
pmvnorm(upper=rep(xl,length(mu)), mean=mu, sigma=sigma)

However on re-reading the question, it seems to be asking the probability that a particular element of the vector is maximal. In this case, I'd agree with G. Jay Kerns.
A: 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).
