The direct way to do this is to integrate $1$ using the spherical volume element $$ d^nV = r^{n-1} \sin^{n-2}(\phi_1) \, \sin^{n-3}(\phi_2) \, \cdots \sin(\phi_{n-2}) \, dr \, d\phi_1 \, \cdots \, d\phi_{n-1}. $$ Here, for $\phi_k$ goes from $0$ to $\pi$ for $k = 1, \dots, n-2$, and $\phi_{n-1}$ goes from $0$ to $2 \, \pi$.

Since $$ \int_0^\pi \sin^k \phi \, d\phi = B\left(\frac12, \frac{k+1}{2} \right) =\frac{ \Gamma\left(\frac{1}{2} \right) \Gamma\left(\frac{k+1}{2}\right) } { \Gamma\left(\frac{k+2}{2} \right) }, $$ where $B(p,q)$ is the beta function, we get $$ \begin{align} \int d^nV &= \int_0^1 r^{n-1} \, dr \, \frac{ \Gamma\left(\frac{1}{2} \right) \Gamma\left(\frac{n-1}{2}\right) } { \Gamma\left(\frac{n}{2} \right) } \frac{ \Gamma\left(\frac{1}{2} \right) \Gamma\left(\frac{n-2}{2}\right) } { \Gamma\left(\frac{n-1}{2} \right) } \cdots \frac{ 2 \, \Gamma\left(\frac{1}{2} \right) \Gamma\left(\frac{1}{2}\right) } { \Gamma\left(\frac{2}{2} \right) } \\ &= \frac{2}{n} \frac{ \left[ \Gamma\left(\frac{1}{2} \right)\right]^n } { \Gamma\left(\frac{n}{2} \right) } = \frac{ {\sqrt\pi}^n } { \Gamma\left(\frac{n}{2} + 1\right) }. \end{align} $$

Another way is to use the formula for the Fourier transform of a spherical function $$ \tilde f(k) = \int f(r) \, e^{i\mathbf k \cdot \mathbf r} \, d\mathbf r = \frac{(2\pi)^{n/2}} { k^{n/2 - 1} } \int_0^\infty f(r) J_{n/2-1}(k\,r)\, r^{n/2} \, dr, $$ where $J_{n/2-1}(k \, r)$ is the Bessel function.

In our case $f(r) = \Theta(1-r)$ is the step function. Then, by using the recurrence relation $$ \frac{d}{dx} \left[ r^{\nu} J_{\nu}(x) \right] = r^{\nu} J_{\nu-1}(x), $$ with $\nu = n/2$, we get $$ \tilde f(k) = \frac{(2\pi)^{n/2}} { k^{n/2} } J_{n/2}(k). $$ For the limit of $k\rightarrow 0$, we get the volume $$ \lim_{k \rightarrow 0} \tilde f(k) = \frac{(2\pi)^{n/2}} { k^{n/2} } \frac{(k/2)^{n/2}}{\Gamma(\frac{n}{2}+1)} = \frac{\pi^{n/2}} {\Gamma(\frac{n}{2}+1)}. $$