I am aware of the geometric argument for computing $\mathbb{P}(X >0, Y>0)$ when $X$ and $Y$ are correlated normal random variables with mean zero, but is there a similar result (approximation?) for a larger number of normal r.v.s? The co-variance matrix that I have in mind is $$\begin{matrix} 3 & 1 & 1\\ 1 & 3 & 1\\ 1 & 1 & 3 \end{matrix} $$ in dimension three and defined similarly in higher dimensions (threes along the diagonal and ones elsewhere).
Letting $X,Y,Z$ have a multivariate normal distribution with the above co-variance matrix $\mathbb{P}(X >0, Y>0, Z >0)$ is equal to $\mathbb{P}(U_1 > U_2, U_1>U_3)$ where $U_1 \sim \mathcal{N}(0,1)$ and $U_2,U_3 \sim \mathcal{N}(0,2)$ (with $U_1,U_2$ and $U_3$ all independent). Perhaps there is a way to exploit this? In the case of $U_1 \sim \mathcal{N}(0,2)$ it is clear that $\mathbb{P}(U_1 > U_2, U_1 > U_3) = 1/3$ since each of $U_1,U_2$ and $U_3$ has equal probability of being the largest of the three.