Let $S^{n-1}\subset\mathbb{R}^n$ be the unit sphere, and $v_1, \cdots, v_m\in S^{n-1}$ be $n$-dimensional unit vectors. Each of these vectors defines an $n-1$ dimensional hyperplane, which cuts $S^{n-1}$ in exactly half. This hyperplane is oriented by the vector $v_i$, meaning it designates a set of exactly half of the points on the sphere; call this set $A_{v_i}$. Now, let $$A=\bigcup_{1\leq i\leq m}A_{v_i}$$
What is the surface area of $A$? Or, equivalently, what portion of the total surface area of $S^{n-1}$ is contained in $A$? More importantly, how does one compute this directly from $v_1, \cdots, v_m$?
Here are a few remarks that may be helpful. First, note that $A$ can be thought of as $A=\{v\in S^{n-1}\text{ | }\exists i\text{ with } v\cdot v_i\geq 0\}$, since the half-$\mathbb{R}^n$ space which each vector $v_i$ defines is precisely equivalent to the set of all vectors which project non-negatively onto $v_i$. Additionally, this question may be easier to answer with low values for $m$; straightforwardly, when $m=2$, one can just look at the angle between the vectors to determine the answer, though I do not see how to extend this… I think that an even answer which treats cases for values of $m\leq 5$, for example, would be interesting.
Edit: It seems this question may be quite a bit harder than anticipated, as can be see here. However, as mentioned above, I still think that an explicit answer for the first few values of $m$ would be interesting. Let $|A|$ denote the measure of $A$. When $m=2$, the answer is simply $|A|=\frac{\pi-\cos^{-1}(v_1\cdot v_2)}{2\pi}$. When $m=3$, the problem comes down to computing the area of the resulting spherical triangle, though I am not sure about the details… I will accept an answer which treats such cases (with $m$ up to 4 or 5, or higher if you can!).
Best Answer
For $n=3$, a spherical polygon's area is given by the Gauss-Bonnet theorem:
$$|A|=2\pi-\sum_{i=1}^m\phi_i$$
$$=\sum_{i=1}^m\theta_i-(m-2)\pi$$
where $\theta_i$ is the interior angle and $\phi_i$ is the exterior angle (both tangent to the sphere) at the $i$th vertex; $\theta_i+\phi_i=\pi$.
Each angle $\phi_i$ is the angle between the planes defining the two edges. In terms of the vectors $a_i\in\mathbb S^2$ representing the vertices, the angles are given by
$$\cos\phi_i=-\cos\theta_i=\frac{-(a_{i-1}\wedge a_i)\cdot(a_i\wedge a_{i+1})}{\lVert a_{i-1}\wedge a_i\rVert\cdot\lVert a_i\wedge a_{i+1}\rVert}=\frac{(a_{i-1}\times a_i)\cdot(a_i\times a_{i+1})}{\lVert a_{i-1}\times a_i\rVert\cdot\lVert a_i\times a_{i+1}\rVert}$$
or in terms of the planes' normal vectors,
$$=v_{i-1}\cdot v_i.$$
But this doesn't tell us whether $\phi_i$ is positive or negative; for that we also need
$$\sin\phi_i=a_i\cdot(v_{i-1}\times v_i)=-I(a_i\wedge v_{i-1}\wedge v_i)=(-Ia_i)(v_{i-1}\wedge v_i)$$
where $I=e_1e_2e_3$. (I'm using the language of geometric algebra.) Likewise, $\sin\phi_i$ alone doesn't tell us whether $\phi_i$ is acute or obtuse. To uniquely determine $\phi_i$, the two formulas can be combined into
$$\exp(Ia_i\phi_i)=\cos\phi_i+Ia_i\sin\phi_i=v_{i-1}v_i.$$
The polygon's perimeter $|\partial A|$ is the sum of the angles $\beta_i$ (cutting through the centre of the sphere) between consecutive vertices. These are given by
$$\exp(-Iv_i\beta_i)=\cos\beta_i-Iv_i\sin\beta_i=a_ia_{i+1}.$$
In your case, $A$ is (the complement of?) a convex polygon, so all $\phi_i$ have the same sign, and thus
$$|A|=2\pi\mp\sum_{i=1}^m\arccos(v_{i-1}\cdot v_i).$$
(Of course we define $v_0=v_m$.) If the set $\{v_i\}$ is unordered so $v_{i-1}$ is unknown, then you can determine whether $v_i$ and $v_j$ are adjacent by whether $v_i\times v_j$ is contained in all the other half-spaces.
For general $n$, I'm thinking of using the divergence theorem to reduce the integral over the polytope to an integral over its boundary. This would require finding a vector field $F$ on the sphere with divergence $1$:
$$|A|=\int_Ad^{n-1}x\overset?=\int_A(\nabla\cdot F)\,d^{n-1}x=\oint_{\partial A}({\bf n}\cdot F)\,d^{n-2}x$$
(${\bf n}$ is the unit vector tangent to the sphere and normal to $\partial A$. For your polytope, this is ${\bf n}=v_i$ on each part of the boundary.)
Using spherical coordinates, here's one example of such a vector field:
$${\bf x}=(\cdots((e_1\cos\theta_1+e_2\sin\theta_1)\cos\theta_2+e_3\sin\theta_2)\cdots\cos\theta_{n-1}+e_n\sin\theta_{n-1})$$
$$F=\theta_1\frac{\partial{\bf x}}{\partial\theta_1}=\theta_1\,{\bf x}\cdot(e_1e_2)$$
We can eliminate the dependence on the coordinate system by noting that $e_1e_2=B$ could be any unit bivector ($B^2=-1$), and that $\theta_1$ is the angle of ${\bf x}$ in the $B$ plane:
$$\exp(\theta_1B)=e_1\frac{({\bf x}\cdot B)B^{-1}}{\lVert{\bf x}\cdot B\rVert}$$
I still don't know whether this can be used to find a more explicit formula for $|A|$...