Here are some exact answers for the one-dimensional case $(d=1)$: $$N=2\negthinspace:\ \frac{1}{2}(1-a^2)$$
$$N=3\negthinspace:\ \frac{3}{4}(1-a^2)$$
$$N=4\negthinspace:\ \frac{1}{8}(1-a^2)(7+a^2)$$
$$N=5\negthinspace:\ \frac{5}{16}(1-a^2)(3+a^2)$$
$$N=6\negthinspace:\ \frac{1}{32}(1-a^2)(31+16a^2+a^4)$$
where $$a=\text{erf}\left(\frac{x-\mu}{\sqrt{2}\sigma}\right)$$
I got these using Mathematica, with Expectation[ Boole[Min[a, b] < x < Max[a, b]], {a [Distributed] NormalDistribution[], b [Distributed] NormalDistribution[]}] // FullSimplify and obvious variants.
Perhaps someone else will see a pattern in the results or extend them to higher dimensions.
Update: Exact formulas for higher dimensions do not look promising.
Consider the toy question: what is the probability that $(1/2, 3)$ lies in the convex hull of $(0,1)$, $(1,2)$, and $(a,b)$, where $a$ and $b$ are both normally distributed and independent? The answer is
which Mathematica does not simplify further. The answer to the original question with $N=3, d=2$ would require four more integrals beyond that.