I worked out the answer for the 2 by 2 case as well.

First, when dealing with 2 by 2 matrices in general, a convenient variable change is:

$$a\rightarrow\frac{w+x}{\sqrt{2}},d\rightarrow\frac{w-x}{\sqrt{2}},c\rightarrow\frac{y-z}{\sqrt{2}},b\rightarrow\frac{y+z}{\sqrt{2}}.$$

Then $a^2+b^2+c^2+d^2 = w^2+x^2+y^2+z^2$. And the determinant $(ad-bc) = \frac{1}{2}(x^2+y^2-w^2-z^2)$.

(Aside: this set of coordinates lets you see for instance that the set of rank 1 matrices in the space of 2D matrices realized as $\mathbb{R}^4$ is a cone over the Clifford torus, since $x^2+y^2 = w^2+z^2$ on a sphere $x^2+y^2+w^2+z^2=r^2$ implies $x^2+y^2 = r^2/2$ and $w^2+z^2 = r^2/2$, which are scaled equations for a flat torus)

Let $r_1^2 = x^2+y^2, r_2^2 = w^2+z^2$. (These are radial coordinates of a coordinate system consisting of two orthogonal 2D cylindrical coordinate systems). Then the norm squared is:

$$\frac{1}{2}\left(r_1^2+r_2^2 + \sqrt{ (r_1^2+r_2^2)^2 - (r_1^2-r_2^2)^2 }\right)$$

When this is less than one, this corresponds to the region plotted below:

Note that each point in the $r_1,r_2$ picture corresponds to a different "torus", $x^2+y^2=r_1^2, w^2+z^2=r_2^2$.

We can now integrate over the shaded in region, $\int_{region} dw dx dy dz$.

This 4-D integral can be reduced to 2D using $r_1$ and $r_2$, since $dx dy = 2\pi r_1 dr_1, dw dz = 2\pi r_2 dr_2$:

$$(4\pi^2) \int_{region} dr_1 dr_2 r_1 r_2. $$

Now, note that we can rewrite $r_2$ in terms of $r_1$. In particular, after some manipulation of our norm, the shaded-in region is defined by $r_2^2 \leq 2-2\sqrt{2}r_1+r_1^2=(\sqrt{2}-r_1)^2$. Hence $r_2\leq \sqrt{2}-r_1$, and we can evaluate the $r_2$ integral:

$$4\pi^2 \int_{r_1=0}^\sqrt{2} dr_1 r_1 \int_{r_2=0}^{\sqrt{2}-r_1} r_2 dr_2 \\
= 4\pi^2 \int_{r_1=0}^\sqrt{2} dr_1 r_1 (\sqrt{2}-r_1)^2/2\\
= (4\pi^2) (1/6).$$

This yields $2\pi^2/3$, as Armin found.