In an n-dimensional Euclidian space, an orthant occupies $\frac{1}{2^n}$ of the volume of the $n$-dimensional sphere. Let's call that a spherical orthant. For example, the "all-positive" spherical orthant is given by
$$x_1 > 0, \qquad x_2 > 0, \qquad \cdots \qquad ,x_n > 0.$$
Let's enclose this (or any other) orthant by a hypersector with colatitude angle $\Theta$, which can be taken from the cone's central axis vector $(1,1,\cdots, 1)$ and one of the bounding vectors $(1,0,0,\cdots , 0)$ to be $\cos \Theta = 1/\sqrt{n}$. How will the volume of this hypersector relate to the volumes of the $n$-dimensional sphere (larger) and the $n$-dimensional spherical orthant (smaller)?
Following the very nice article by S. Li here, the solution requires regularized incomplete beta functions $I_{\sin^2 \Theta} (\frac{n-1}{2}, \frac{1}{2})$.
We have that
$$
V_{orthant} / V_{sphere} = \frac{1}{2^n}
$$
and
$$
V_{sector} / V_{sphere} = \frac12 I_{1-1/n} (\frac{n-1}{2}, \frac{1}{2}) \qquad .
$$
The interesting question is: how does the last result behave for large $n$? And what is the large-n-behavior of
$$
V_{sector} / V_{orthant} = 2^{n-1} I_{1-1/n} (\frac{n-1}{2}, \frac{1}{2}) \qquad ?
$$The large-n-behavior is sought as a function of $n$, which also can be approximated or just bounded. This may be taken from the large-n-behavior of the
given regularized incomplete beta function, or from other (analytic-)geometrical considerations.
We can flip the regularized incomplete beta function
$$I_{1-\frac1n}\left(\frac{n-1}2,\frac12\right) = 1 - I_{\frac1n}\left(\frac12,\frac{n-1}2\right).$$
And then use the expansion: $$ I_z(a,b) = \frac{z^a}{B(a,b)}\sum_{k=0}^\infty\frac{(1-b)_kz^k}{(a+k)k!}. $$
We can notice that $z\propto 1/n$ and $b\propto n$, so the sum will go to some constant as $n\to\infty$, so what we are really interested in is $z^a/B(a,b)$.
$$ \log B\left(\frac12,\frac{n-1}2\right) = \log \Gamma(\frac12) + \log \Gamma(\frac{n-1}2) - \log \Gamma(\frac n2) \approx \\ \frac{n-1}2\log\frac{n-1}2 - \frac{n-1}2 - \frac{n}2\log\frac{n}2 + \frac{n}2\approx -\frac12\log n $$
So $B(\frac12\,\frac{n-1}2)\approx 1/\sqrt n$, but $z^a$ is also $1/\sqrt n$. Thus $\lim_{n\to\infty}I_{\frac1n}\left(\frac12,\frac{n-1}2\right)=C=\mathrm{const}$ (checking this with wolfram shows $C=0.682689\ldots$). Thus the volume ratio behaves just like an exponential function $2^n$.