As a general remark: Circle, Cube, whatever does not matter - the unbounded problem gives the same numbers.

4 3-dimensional cuts can produce 16 4-dimensional volumes. The easiest way to see this is to use the "coordinate cubes" as cuts: x_1=0, x_2=0, x_3=0, x_4=0

Defining the numbers for the n-dimensional problem with i cuts as f(n,i), I think that the formula f(n,i)=f(n,i-1)+f(n-1,i-1) (with the special case n=3 explained at OEIS) should hold.

f(n,0)=1

f(1,i)=i+1 (these are i dots on a line)

f(2,i)=1+i(i+1)/2 and the formula fits: f(2,i) = f(2,i-1)+f(1,i-1)=1+(i-1)(i)/2+(i-1)+1=1+i(i+1)/2

Another interesting thing: Formulas like these are usually polynomials, here with the dimension of the cutted volume as order. As the value for i<=n cuts is always 2^i (use the coordinates as cuts), the n+1 trivial cases are sufficient to determine all n+1 parameters of the polynomial.