Let $Q\subset \mathbb{R}^3$ denote the cube $[-1,1]\times[-1,1]\times[-1,1]$.
Let $\xi\in\mathbb{S}^2$ be a vector on the unit sphere centered at the origin.
Let $\xi^{\perp}$ denote the linear subspace that is orthogonal to the vector $\xi$.
Denote $Q\mid_{\xi^{\perp}}$ as the projection of $Q$ onto the plane $\xi^{\perp}$.
Question: What shape is $\bigcap_{\xi\in\mathbb{S}^2}Q\mid_{\xi^{\perp}}$?
I don't think you've "mathematized" this correctly. Without even bothering with projecting $Q$, isn't the intersection of all the $\xi^{\perp}$ just the origin?
So you probably want to consider all $SO(3)$ rotations of the cube, and project them all onto a fixed plane, say the $x−y$ plane.
Once you've fixed this issue, the resulting intersection will be rotation invariant, as any element of $SO(3)$ can be composed with a suitable rotation around the $z$-axis. Additionally, rotating and projecting a convex set yields a convex set, and intersecting a bunch of convex sets yields a convex set. That doesn't seem to leave a lot of possibilities (hint, hint).
If you already knew all that, and were just curious about the resulting radius, take a look at the largest sphere you can inscribe in your cube.