Since you speak of an "easy way" to compute the volume in this third case, I can only say this: I don't think that there is an easy way. My thoughts and (modest) initial efforts to solve this follow, but they don't finish with an actual formula, so read on only for inspiration, not a solution.
First, fix the sphere radius $r$ at $1$. You'll get an answer $U(L)$ in terms of $L$, but for any $(L,r)$ pair, the answer will just be
$$
Vol(L, R) = r^3 U(L/r)
$$
so that simpler answer will implicitly provide the more complex one
The "interesting case" is when the corner of the cube "pokes out" of the sphere, but any two adjacent "poking out" bits don't poke out far enough to reveal the entire edge between them on the outside of the sphere.
To address this case, I want to rotate the whole problem, so that one vertex of the cube lies at $P = c(0,1,0)$, where $c = \sqrt{3}L/2$, and one edge of the cube lies in the $yz$ plane with $z > 0$, i.e., $\theta = 0$ in polar coordinates (where $y = 1$ is the north pole, where $\phi = 0$, and at $(0, 0, 1)$ we have $\theta = 0, \phi = \pi/2$, just to establish the coordinates I'll be using.
Now looking down on the north pole, we see the three edges of the protruding corner of the cube; in polar coordinates, they lie along the planes $\theta = 0, \theta = 2\pi/3, \theta = 4\pi/3$. I'm going to compute the volume $V$ of one of the three "wedges" of this protruding piece.
The volume of the whole protrusion will be $3V$, and the volume of all protrusions will $24V$, so the total cube-sphere intersection will be the cube volume $L^3$, with these protrusion subtracted, i.e.,
$$
U(L) = L^3 - 24 V.
$$
So what is $V$? It's an integral, in polar coordinates, of the radius-function $h(\theta, \phi)$ for the cube minus the radius function for the sphere, which is just the constant $1$. The limits of the integral are
$$
0 \le \theta \le 2\pi/3
$$
and
$$
0 \le \phi \le ???
$$
To figure out that missing bound, I'm going to write the equation, in polar coordinates, for the cube-face that lies between $\theta = 0$ and $\theta = 2\pi/3$, i.e., the face on which $x$ is always positive, and which contains $c(0,1,0)$.
The unit normal to this plane is
$$
n = \begin{bmatrix}
\frac{1}{\sqrt{2}} \\
\frac{1}{\sqrt{3}} \\
\frac{1}{\sqrt{6}}
\end{bmatrix}
$$
The plane equation is then
$$
X \cdot n = L,
$$
When you convert this to polar coordinates, you get a rather messy expression for the radius of the vector $X$, which is what I've called $h(\theta, phi)$. Setting this messy expression to $r = 1$ determines the curve of intersection between the cube-face and the sphere, which you can solve for $\phi$ in terms of $\theta$, getting some expression
$$
\phi = q(\theta)
$$
And now
$$
V = \int_{\theta=0}^{\frac{2\pi}{3}} \int_{\phi = 0}^{q(\theta)} s ~d\theta.
$$
$$
V = \int_{\theta =0}^{\frac{2\pi}{3}}\int_{\phi = 0}^{q(\theta)} (h(\theta, \phi) - 1) J(\theta, \phi) ~d\phi ~d\theta.
$$
where $J$ is the appropriate Jacobian ($\sin \phi ? \cos \phi ?$) for this set of polar coordinates.
