You don't want to do the integral directly: hence the hint. One way to integrate is to split
$$ e^{-|x|^2} = e^{-(x_1^2+x_2^2+\dotsb+x_n^2)} = e^{-x_1^2} e^{-x_2^2} \dotsm e^{-x_n^2} , $$
which is a product of $n$ one-dimensional integrals, all of which we know evaluate to
$$ \int_{-\infty}^{\infty} e^{-x^2} \, dx = \sqrt{\pi}. $$
On the other hand, if you substitute in spherical coordinates, you have $e^{-|x|^2}=e^{-r^2}$, and the volume element is
$$ r^{n-1}\sin^{n-2}\phi_1 \sin^{n-3}\phi_2\cdots\sin \phi_{n-2} \, dr \, d\theta \, d\phi_1\cdots d\phi_{n-2}, $$
which means that the integral splits this time as:
$$ \left( \int_0^{\infty} r^{n-1} e^{-r^2} \, dr \right) \left( \int_R \sin^{n-2}\phi_1 \sin^{n-3}\phi_2\cdots\sin \phi_{n-2} \, d\theta \, d\phi_1\cdots d\phi_{n-2} \right). $$
The second integral is the one you want, and you can do the first using the Gamma function, which will give you the answer you want when you divide: that is, you have two different evaluations of the same quantity, so they must be equal,
$$ \pi^{n/2} = \int_{\mathbb{R}^n} e^{-|x|^2} \, dx = \left( \int_0^{\infty} r^{n-1} e^{-r^2} \, dr \right) \left( \int_R \sin^{n-2}\phi_1 \sin^{n-3}\phi_2\cdots\sin \phi_{n-2} \, d\theta \, d\phi_1\cdots d\phi_{n-2} \right), $$
so
$$ \int_R \sin^{n-2}\phi_1 \sin^{n-3}\phi_2\cdots\sin \phi_{n-2} \, d\theta \, d\phi_1\cdots d\phi_{n-2} = \frac{\pi^{n/2}}{\int_0^{\infty} r^{n-1} e^{-r^2} \, dr}. $$
