Calculate $\Phi_n((-\infty)_n, 0_n; 0_n, I_n + \rho e_n e_n')$, where $\Phi_n((a)_n, (b)_n; \mu, \Sigma) = P(X_i \in (a_i, b_i) \forall i=1,2\dots n)$ for $X$ n-variate normal with mean $\mu$ and covariance matrix $\Sigma$. In this notation, $I_n$ is the $n$-dimensional identity matrix and $e_n$ is the vector of length $n$ containing 1 as all elements, implying $e_ne_n$ is the matrix of size $n\times n$ with 1 as all elements.
There is plenty of literature on finding this for $n\leq 4$, for a general $\Sigma$, but I am instead interested in the problem, where all covariances are equal to each other. I am aware that I cannot in general expect to find an analytical solution, but in the case $\rho = 1$, I have calculated it for $n=1, 2, 3$ to be equal to $\frac{1}{n+1}$, and a simulation for greater $n$ suggests that it might be a general result.
To find the analytical solution, I have used formula (5.2) from here, saying $p_n(\tau) = \frac{1}{2^n}\sum_{k=0}^{\lfloor\frac{n}{2}\rfloor}\frac{A_n^{2k}}{\pi^k}F_{n,k}(\tau)$, where $F_{n,0}(\tau) = 1$ for any $n$, $F_{n,1}(\tau) = \sin^{-1}(\tau)$ for $n$ greater than 2 and $F_{n,k}(\tau) = \int_0^\tau F_{n-2, k-1}(\frac{x}{1+2x})d\sin^{-1}x$. Also $A^{2k}_n =\frac{n!}{(n-2k)!}$. $P_n(\tau) = \Phi_n((-\infty)_n, 0_n; 0_n, I_n + \rho e_n e_n')$ for $\tau = \frac{\rho}{1+\rho}$.
At the moment, I need the results for $\rho = \frac{1}{2}, \frac{2}{3}, 1$, equivalent to $\tau = \frac{1}{3}, \frac{2}{5}, \frac{1}{2}$ and would be a great help for my present project, but the more general your response is, the more future readers it can help.
My first idea to calculate $\Phi_n((-\infty)_n, 0_n; 0_n, I_n + \rho e_n e_n')$ would be to calculate it for the first cases and use a proof by induction using the result I have introduced earlier, but as it does not require only the value of interest, but it requires the integral over all values lower ratios $\tau$, and a typical proof by induction does not seems to work.
I look forward to hearing your responses! Thank you very much for the time spent on this.
I hereby link to one of my previous questions, where there is a good answer that gives some equivalent expressions for the orthant probability. This might be helpful in potentially gaining a solution in these cases.