Intersection of squares/cubes/hypercubes. 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!):
        

 A: 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
A: 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.
