The volume of the delta-ball of the special orthogonal group can be computed exactly by applying the Weyl integration formula: (Without loss of generality, we assume that the delta-ball is around the unit group element).
a. One notices (Again due to the invariance under the Haar measure) that the characteristic function of the delta ball is a class function. Thus upon the application of the Weyl integration formula we are left only with the radial part on the eigenvalues which is a floor(N/2) dimensional $\lfloor N/2\rfloor$-dimensional integral for SO(N). $\mathrm{SO}(N)$. Here, the radial integral is described explicitely.
b. The eigenvalues of an orthogonal matrix of dimension N=2m+1 $N=2m+1$ are 1 $1$ and m $m$ pairs exp(i*phi_ m) $\exp(i \phi_ m)$ and exp(-i*phi_ m)$\exp(-i \phi_ m)$, 0<=phi_ $0\leq\phi_ 1 <= . . . <= phi_m <=pi. \leq\ldots\leq\phi_m \leq\pi$. In the case of even dimensions, the unit eigenvalue is absent.
c. The delta-ball condition on the eigenvalues becomes:
$$ |exp(i*phi_ k)-1|<=delta, \exp(i\phi_k)-1|\leq\delta , $$ which implies: phi_ k<=2 * arcsin(sqrt(delta/2))$$\phi_k\leq 2 \arcsin\sqrt{\delta/2}.$$
d. Applying the Weyl integration formula, we obtain for the odd case SO(2*m+1):
Vol(delta-ball) $\mathrm{SO}(2m+1)$:
$$ \mathrm{Vol}(\delta\mathrm{-ball}) = 2^(m^2)/(pi^m * m!)* int_ phi_ 1<=phi_ \frac{2^{m^2}}{\pi^m m!} \int_{\phi_1\leq\ldots\leq\phi_m \leq 2 <= . . . <=phi_ m <=2 * arcsin(sqrt(delta/2))*
prod_ 1 < = \arcsin\sqrt{\delta/2}} \prod_{1\leq j < k < = \leq m} (cos(phi_ k)-cos(phi_ j)^2 prod_ l sin^2(phi_ l) dphi_ 1 . . . dphi_ k\cos\phi_k-\cos\phi_j)^2 \prod_l \sin^2(\phi_l) d\phi_1 \cdots d\phi_k. $$
e. For the even dimensional case, the only changes are 2^(m^2) $2^{m^2}$ is replaced 2^((m-1)^2) $2^{(m-1)^2}$ and the sine terms are absent.
