Ok, i've done some calculating:
3x3 magic squares have a base of three elements (eg, you can choose 3 numbers in the sqaure, and then the rest is fixed).
4x4 magic squares have a base of 8 elements.
If i haven't made a mistake, 4x4x4 magic cubes have a 18-dimensional base. I said 15 before, but i forgot that several equations are abundant.
I can show you how to make a base for whatever size magic cube/square you want. It involves heavy calculation though
I'll use magic squares for example (sides of length three). I'll use the following natation: (1, 2, 3, 4, 5, 6,7,8,9) will be a square with a 1 in the top left, 3 in the top right, and a 7 in the bottom left.
Say you want the sums of the sides to be X. You can choose the first two spaces freely, say a and b and choose the 4th space to be c (Not the third, because that one is already set because of your choice of a and b).
The third coordinate is set, namely X-a-b. You see that now the 7th coordinate is set as well: X-a-c. Now the 5th coordinate is set (because the diagonal also has to have sum X) : -X +2a+b+c. You can now complete the rest of the square (using horizontals and verticals) wich gives you the following square (draw it for a clear idea):
(a, b, X-a-b, c, -X+2a+b+c, 2X-2a-b-2c, x-a-c, 2X-2a-2b-c, -2X+3a+2b+2c)
The only sum we haven't checked yet is the first diagonal. If we add up the numbers on the diagonal (1st, 5th and 9th element) we get -3X+6a+3b+3c. To be a magic square this needs to equal X.
X = -3X+6a+3b+3c => X = 1/4 * (6a +3b +3c)
Now we're ready to make the base-elements. We're going to find squares A,B and C so that a*A + b*B + c*C is the square we have now (basically, we seperate our square in a square with b=c=0, a square with a=c=0 and a square with a=b=0).
You'll see quickly that they are the following squares (Note that X is determined by a,b and c!)
A = (1, 0, 1/2, 0, 1/2, 1, 1/2, 1, 0)
B= (0, 1, -1/4, 0, 1/4, 1/2, 3/4, -1/2, 1/2)
C= (0, 0, 3/4, 1, 1/4, -1/2, -1/4, 1/2, 1/2)
(You can check it yourself: a*A + b*B + c*C gives our square with sums X = 1/4*(6a+3b+3c) )
ANY magic 3x3 square is of the form a*A + b*B + c*C with a, b and c anything you want.
This method can be used for 4x4x4 magic cubes as well, or even higher dimensional things with longer sides. The problem is that you'll lose oversight on what is set by your choices. I think you'll have 18 free choices in your magic cube, then the rest is set.
PS There ARE 2x2 magic squares and 3x3x3 magic cubes! They simply aren't interesting because all the numbers are the same)