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->(w+x)/\sqrt{2},d->(w-x)/\sqrt{2},c->(y-z)/\sqrt{2},b->(y+z)/\sqrt{2}.
$$a\rightarrow\frac{w+x}{\sqrt{2}},d\rightarrow\frac{w-x}{\sqrt{2}},c\rightarrow\frac{y-z}{\sqrt{2}},b\rightarrow\frac{y+z}{\sqrt{2}}.$$
Then a^2+b^2+c^2+d^2 = w^2+x^2+y^2+z^2$a^2+b^2+c^2+d^2 = w^2+x^2+y^2+z^2$. And the determinant (ad-bc) = (1/2)*(x^2+y^2-w^2-z^2)$(ad-bc) = \frac{1}{2}(x^2+y^2-w^2-z^2)$.
(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 R^4$\mathbb{R}^4$ is a cone over the Clifford torus, since x^2+y^2 = w^2+z^2$x^2+y^2 = w^2+z^2$ on a sphere x^2+y^2+w^2+z^2=r^2$x^2+y^2+w^2+z^2=r^2$ implies x^2+y^2 = r^2/2$x^2+y^2 = r^2/2$ and w^2+z^2 = r^2/2$w^2+z^2 = r^2/2$, which are scaled equations for a flat torus)
Let r1^2 = x^2+y^2, r2^2 = w^2+z^2$r_1^2 = x^2+y^2, r_2^2 = w^2+z^2$. (These are radial coordinates of a coordinate system consisting of two orthogonal 2D cylindrical coordinate systems filling out 4-space). Then the norm squared is:
(1/2)*(r1^2+r2^2 + \sqrt{ (r1^2+r2^2)^2 - (r1^2-r2^2)^2 })
$$\frac{1}{2}\left(r_1^2+r_2^2 + \sqrt{ (r_1^2+r_2^2)^2 - (r_1^2-r_2^2)^2 }\right)$$
When this is less than one, this corresponds to the region plotted below:
Note that each point in the r1,r2$r_1,r_2$ picture corresponds to a different "torus", x^2+y^2=r1^2, w^2+z^2=r2^2$x^2+y^2=r_1^2, w^2+z^2=r_2^2$.
We can now integrate over the shaded in region, \int_{region} dw dx dy dz$\int_{region} dw dx dy dz$.
This 4-D integral can be reduced to 2D using r1$r_1$ and r2$r_2$, since dx dy = 2π r1 dr1, dw dz = 2π r2 dr2$dx dy = 2\pi r_1 dr_1, dw dz = 2\pi r_2 dr_2$:
(4π^2) \int_{region} dr1 dr2 r1 r2
$$(4\pi^2) \int_{region} dr_1 dr_2 r_1 r_2. $$
Now, note that we can rewrite r2$r_2$ in terms of r1$r_1$. In particular, after some manipulation of our norm, the shaded in-in region is defined by r2^2 ≤ 2-2\sqrt{2}r1+r1^2=(\sqrt{2}-r1)^2$r_2^2 \leq 2-2\sqrt{2}r_1+r_1^2=(\sqrt{2}-r_1)^2$. Hence r2≤ \sqrt{2}-r1$r_2\leq \sqrt{2}-r_1$, and we can evaluate the r2$r_2$ integral:
(4π^2) \int_{r1=0}^\sqrt{2} dr1 r1 \int_{r2=0}^{\sqrt{2}-r1} r2 dr2
= (4π^2) \int_{r1=0}^\sqrt{2} dr1 r1 (\sqrt{2}-r1})^2/2
= (4π^2) (1/6)
$$4\pi^2 \int_{r_1=0}^\sqrt{2} dr_1 r_1 \int_{r_2=0}^{\sqrt{2}-r_1} r_2 dr_2 \\ = 4\pi^2 \int_{r_1=0}^\sqrt{2} dr_1 r_1 (\sqrt{2}-r_1)^2/2\\ = (4\pi^2) (1/6).$$
This yields 2π^2/3$2\pi^2/3$, as Armin found.
