You can find a fully worked-out derivation in these lecture notes. The formula you are looking for is equation (404), written in terms of the Wigner (small)-$d$ matrix. The relationship to the (large)-$D$ matrix goes via the Euler angle parameterization, $$D^{j}_{mm'}(\psi,\theta,\phi)=e^{-im\psi-im'\phi}d^{j}_{mm'}(\theta),$$ The integration over SU(2) with the Haar measure is given in terms of Euler angles by $$U(\psi,\theta,\phi)=\exp(-i(\psi/2)\sigma_1)\exp(-i(\theta/2)\sigma_2)\exp(-i(\phi/2)\sigma_3,$$ $$\int_{\rm{SU}(2)}f(U)\,dU=\frac{1}{16\pi^2}\int_0^{2\pi}d\psi\int_0^\pi\sin\theta d\theta\int_0^{4\pi}d\phi\, f[U(\psi,\theta,\phi)]$$
So the desired integral over a product of three $D$-matrices vanishes unless $m_1+m_2+m_3=0=n_1+n_2+n_3$. In that case the integrations over $\psi$ and $\phi$ give a factor $8\pi$, what remains is the integration over $\theta$. Eq. 404 in the lecture notes shows how that is related to the product of $3j$-symbols, $$\frac{1}{2}\int_0^\pi d\theta\,d^{j_1}_{m_1n_1}d^{j_2}_{m_2n_2}d^{j_3}_{m_3n_3}=\begin{pmatrix}j_{1} & j_{2} & j_{3}\\m_{1} & m_{2} & m_{3}\end{pmatrix}\begin{pmatrix}j_{1} & j_{2} & j_{3}\\n_{1} & n_{2} & n_{3}\end{pmatrix}.$$ Both sides of the equation are equal to zero unless $\vert j_{2}-j_{3}\vert\leq j_{1}\leq j_{2}+j_{3}$.