Intersection of squares/cubes/hypercubes.

Question

One can form a polygon of $4 n$ sides by intersecting $n$ congruent squares (treated as closed sets, i.e., filled squares):
        

Q1. For which of the $k=3,4,\ldots,4n$ can the intersection of $n$ congruent squares result in a $k$-gon? Perhaps not all can be achieved?

Q2. Can the intersection of $n$ congruent cubes result in a polyhedron of $6 n$ faces? I believe so, but an explicit construction would be useful.

Q3. For which $k$ can $k$-face polyhedra result from the intersection of $n$ congruent cubes?

Q4. The questions extend to $\mathbb{R}^d$.

Question Q2 in particular occurred to me as a possible exercise to build 3D intuition.

Added. Here are two cubes intersected to produce a polyhedron of 12 faces (although one can hardly verify that from this single, not well-lighted image!):
        

Answer 1

Any configuration of regular polytopes, where no two polytopes have a pair of parallell planes, and every plane is symmetric wrt. to the configuration, and the intersection is nonempty, should give the max faces.

For Q2 and n odd, we can rotate each cube around an internal diagonal like this

Answer 2

Question 2 :

Even though OP asks about cubes I would like to comment some about squares. There is $n$ squares of side length $2$ s.t. all centers of squares $A_i,\ 1\leq i\leq n $ is located at the origin in $\mathbb{R}^2$.

If $v_i$ is a unit vector in any one side of $A_i$, then assume that the angle between $x$-axis and $v_i$ is strictly larger than $0$ and strictly smaller than $\frac{\pi}{2}$.

Note that each $v_i$ determines the $i$-th square uniquely. Further if $$\sharp\{v_1,\cdots , v_n\}=n,$$ then side corresponded to $v_i$ survives in the intersection of $n$ squares. Hence if we choose $n$ different vectors $v_i$ with that angle condition, then we have $4n$-polygon.

Now we will consider about $n$ cubes :

We will use the similar idea. If $C_i$ are cubes with side length $2$ and origin center, then we have orthonormal frame $\{a_i,b_i,c_i\}$ s.t. each vectors are in sides, and then $a_ib_ic_i$ is an equilateral spherical triangle of side length $\frac{\pi}{2}$.

If $x$ is a point in $\mathbb{S}^2$, then there is $r$ s.t. geodesic sphere $S(r,x)$ in $\mathbb{S}^2$ contains $a_i,\ b_i,\ c_i$ for all $i$. If $y\in S(r,x)$, then we are sufficient to take $a_i$ sufficiently close to $y$, differently, which completes the proof.