Let $v$ be a vector on the unit sphere in $\mathbb{R}^n$ and let $S(\epsilon)$ be the set of vectors $s$ on the same sphere such that $$ |s \cdot v| \leq \epsilon.$$ What is the surface area of $S(\epsilon)$?
If an exact answer is impossible, I'd like to know about an approximate answer, i.e., what sort of function is ${\rm area}(S(\epsilon))$? How fast does it grow?
On
This quantity is much studied in high-dimensional convex geometry; an excellent starting point is Keith Ball's An Elementary Introduction to Modern Convex Geometry (MSRI, 1997), especially Lecture 2. See also this question.
The unit sphere in $\mathbb{R}^n$ is denoted by $\mathcal{S}^{n-1}$. Its surface area is given by $S_{n-1}$ on Wikipedia, but let's call it $A_n$ here, because, well, we're already using $S$ a lot, and the $n-1$ is confusing: $$A_n = 2\pi^{n/2} / \Gamma(n/2).$$ Now let's look at a different set than yours: $$C(\epsilon) = \left\{s\in\mathcal{S}^{n-1}\,|\,s\cdot v>\epsilon\right\}$$ Assume that $0\leq\epsilon\leq 1$. This is a hyperspherical cap. The height of this cap is actually just $1-\epsilon$, and using the formula in the linked article, we see that $$A_{n,\epsilon} = \tfrac{1}{2} A_n (1-G_n(\epsilon)/G_n(1)) \qquad G_n(q) \triangleq \int_0^q (1-t^2)^{(n-1)/2}\, dt.$$ Now, your set $S(\epsilon)$ is in fact the surface of the sphere with two opposite hyperspherical caps removed. Therefore, its area is $$A = A_n - 2A_{n,\epsilon} = A_n G_n(\epsilon)/G_n(1) = \frac{ 2\pi^{n/2} G_n(\epsilon) }{ \Gamma(n/2) G_n(1) }.$$ I've got to say, that's simpler than I thought it would be. In particular, the ratio of the area of $S(\epsilon)$ to the surface area of the entire sphere is just $$\frac{A}{A_n} = \frac{G_n(\epsilon)}{G_n(1)}.$$