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}.

Then 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).

(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 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 r1^2 = x^2+y^2, r2^2 = w^2+z^2 be the radial coordinates of two 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 })

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

![spectral norm ball][1]


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

But this 4-D integral can be reduced to 2D using r1 and r2, since dxdy = 2π r1 dr1, dw dz=2π r2 dr2, and furthermore we can write r2 in terms of r1:

    (4π^2) \int_{region} dr1 dr2 r1 r2 

The region can be defined by r2^2 ≤ 2-2\sqrt{2}r1+r1^2=(\sqrt{2}-r1)^2.  Hence r2≤ \sqrt{2}-r1, and we can evaluate the r2 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)

This then yields 2π^2/3, as Armin found.

  [1]: http://i583.photobucket.com/albums/ss275/jaspercrowne/spectralnormball.png