"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...

No, 14 is not the only possibility.

For example: Arrange the dice, so that you only see 1,2 and 3 pips and all the 2's are on the upper and lower face of the big cube. This gives you face sum 8.

Please ask your other questions as separate questions if you are still interested.

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 $28-k$.