"8 Dice arranged as a Cube" Face-Sum Equals 14 Problem I found this here:

Sum Problem
Given eight dice. Build a $2\times 2\times2$ cube, so that the sum of the points on each side is the same.
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Here is one of  20 736 solutions with the sum 14.
  You find more at the German magazine "Bild der Wissenschaft 3-1980". 

Now my question:
Is $14$ the only possible face sum? At least, in the example given, it seems to related to the fact, that on every face two dice-pairs show up, having $n$ and $7-n$ pips. Is this necessary? Sufficient it is...

A note regarding your first question: If a solution with face sum $k$ exists, then a solution with face sum $28-k$ also exists. To see this, start with a solution $S$.  Note that three sides of each die are exposed. If we move each die to the position exactly opposite where it is in $S$, this creates an arrangement of dice $S^\prime$. Now consider the front face of $S$ and the back face of $S^\prime$. Together these contain four pairs of opposing faces of dice, and each opposing pair sums to 7. So the front face of $S$ and the back face of $S^\prime$, together, have eight numbers that sum to 28; if the front face sums to $k$ then the back face sums to $28-k$.
