The left-invariant metric on $SU(2)$ is defined as $ds^2 = -2 \operatorname{Tr}( g^{-1} \mathrm{d} g \cdot g^{-1} \mathrm{d} g )$.
Let us choose the following group element parameterization:
$$
g(\theta, \psi, \phi) = \exp( \psi \tau_3 ) \exp( \theta \tau_2 ) \exp( \phi \tau_3 )
$$
where
$$
\tau_1 = \left(
\begin{array}{cc}
0 & -\frac{i}{2} \\
-\frac{i}{2} & 0 \\
\end{array}
\right) \quad\quad
\tau_2 = \left(
\begin{array}{cc}
0 & -\frac{1}{2} \\
\frac{1}{2} & 0 \\
\end{array}
\right) \quad\quad
\tau_3 = \left(
\begin{array}{cc}
-\frac{i}{2} & 0 \\
0 & \frac{i}{2} \\
\end{array}
\right)
$$
Matrices $\tau_i$ are realizations of $\mathfrak{su}(2)$ Lie algebra, i.e. $\left[ \tau_1, \tau_2 \right] = \tau_3$ and cyclic variants of that. With this representation
$$
g(\theta, \psi, \phi) = \left(
\begin{array}{cc}
\cos \left(\frac{\theta }{2}\right) e^{-\frac{1}{2} i (\psi +\phi )} & \sin
\left(\frac{\theta }{2}\right) \left(-e^{\frac{1}{2} i (\phi -\psi )}\right) \\
\sin \left(\frac{\theta }{2}\right) e^{-\frac{1}{2} i (\phi -\psi )} & \cos
\left(\frac{\theta }{2}\right) e^{\frac{1}{2} i (\psi +\phi )} \\
\end{array}
\right)
$$
Here $0 \le \theta 2\pi$, $0 \le \phi, \psi < 4 \pi$. Some algebra leads to
$$
\mathrm{d}s^2 = \mathrm{d} \theta^2 + \mathrm{d} \phi^2 + \mathrm{d} \psi^2 + 2 \cos \theta \mathrm{d} \psi \mathrm{d} \psi = h_{i,j} \mathrm{d}x^i \otimes \mathrm{d}x^j
$$
Then the volume element is $\mathrm{d}V = \sqrt{\det h} \cdot \mathrm{d} \theta \, \mathrm{d}\psi \, \mathrm{d}\psi = \vert \sin\theta \vert \cdot \mathrm{d} \theta \, \mathrm{d}\psi \, \mathrm{d}\psi$. The volume of $SU(2)$ then is
$$
\int_0^{2 \pi} \mathrm{d} \theta \int_0^{4 \pi} \mathrm{d} \phi \int_0^{4 \pi} \mathrm{d} \psi \cdot \vert \sin \theta \vert = (4 \pi)^2
$$
We can now repeat this same process using
$$
g^{-4} \cdot \mathrm{d} g^4 = \sum_{k=0}^3 g^{-k} ( g \cdot \mathrm{d} g) g^k
$$
The algebra gets quite more involved, with
$$
\sqrt{ \det \tilde{h} } = 8 \vert \sin \left(\frac{\theta }{2}\right) \cos ^3\left(\frac{\theta }{2}\right) \cos
^2 \left(\frac{\psi +\phi }{2}\right) k(\theta, \phi, \psi) \vert
$$
where $k(\theta, \phi, \psi) = (\cos (\theta -\psi -\phi )+\cos (\theta
+\psi +\phi )+2 \cos (\theta )+2 \cos (\psi +\phi )-2)^2$.
Carrying out the integration
$$
\int_0^{2 \pi} \mathrm{d} \theta \int_0^{4 \pi} \mathrm{d} \phi \int_0^{4 \pi} \mathrm{d} \psi \cdot \sqrt{ \det \tilde{h}} = 64 \pi^2 = ( 8 \pi )^2
$$
As the consequence, according to your formula,
$$
\deg S = \frac{(8 \pi)^2}{(4 \pi)^2} = 4
$$
So it turned out an heavy lifting exercise which confirm intuition behind Ryan's answer in comments.