Let us equip the special orthogonal group $SO(n)$ with a normalized Haar measure $\theta_n$ and let $G_r$ be the subset of rotations $\Omega$ which differ from the identity by (sufficiently small) $r$ in the sense that $|\Omega x - x| \leq r$ for all $x\in S^{n-1}$. What is $\theta_n(G_r)$? Is it of order $r^{c(n)}$ for some $c(n)$?
When $n=2$, it can be seen that $SO(2)$ is isomorphic to the circle group $S^1$ and the normalized Haar measure is just the normalized arclenth measure for $S^1$. So we have $\theta_2(G_r) \sim r$ if I understood it correctly. Perhaps, this might be an exercise problem for $n\geq 3$ as well, but I don't have enough background to compute this explicitly.
There's an estimate for a measure of certain rotations, which seems to be related to the question above, in the proof of the wave packet decomposition in Tao's paper "Endpoint bilinear restriction theorems for the cone, and some sharp null form estimates" (http://arxiv.org/abs/math/9909066v2), which I would like to understand. The estimate is at the end of the proof at 41page of the arxiv paper in case you are interested.
Added
I figured out an upper bound for $\theta_n(G_r)$ that $\theta_n(G_r) \leq C r^{n-1}$ for some constant $C>0$ for all sufficiently small $r$. This may follow from the following identity which connects the Haar measure $\theta_n$ on $SO(n)$ and the surface measure $\sigma_{n-1}$ on $S^{n-1}$.
Let $s \in S^{n-1}$, $B \subset S^{n-1}$, and define the map $f:SO(n) \to S^{n-1}$ by $f(\Omega) = \Omega s$. Then $\sigma_{n-1}(B) = \theta_n(f^{-1}(B))$.
A proof of the statement can be found in Geometric Integration Theory by Krantz and Parks (Actually their statement (Proposition 3.2.1) is stronger than that I stated above. But I guess that there is a small gap in the proof when they assume $f^{-1}(B) = A$.) For the upper bound, let $A\subset SO(n)$. Then as a corollary, one has $\theta_n(A) \leq \theta_n(f^{-1}(f(A))) = \sigma_{n-1}(f(A))$. Thus the upper bound follows by observing that $f(G_r)$ is a cap of radius $r$ on $S^{n-1}$. But this does not seem to answer the lower bound question.