We will say that a die of the large cube is of type I, type II or type III according to the number of its faces which are visible (e.g. a die sharing a vertex with the large die is a type III die). Every large cube contains $6$ type I, $12$ type II and $8$ type III dice.

The minimum sum is obtained when each type I die shows $1$, each type II die shows $1$ and $2$ and each type III die shows $1$, $2$ and $3$. Its value is $S_{\min} = 6 \cdot 1 + 12 \cdot 3 + 8 \cdot 6 = 90$.

The maximum sum is obtained when each type I die shows $6$, each type II die shows $5$ and $6$ and each type III die shows $4$, $5$ and $6$. Its value is $S_{\max} = 6 \cdot 6 + 12 \cdot 11 + 8 \cdot 15 = 288$.

We will show that the sum can be any number from $90$ to $288$. In order to do this we will start from the minimum sum and we will rotate the dice so that the sum increases by $1$ at each step.

Rotating a type I dice we can increase the total by $1$ each time, until this die shows $6$. This way we can make each type I die show a $6$.

We rotate now a type II die from $1+2$ to $5+6$ and, in the same time, we rotate two of the type I dice to $1$ and $4$ respectively; this way the sum increases by $1$. We then rotate back the type I dice, little by little, so they show $6$ again, increasing the sum each time by $1$. We repeat this move until each type II die shows $5+6$.

Then we rotate a type III die so it shows $4+5+6$, rotating in the same time two type I dice so they show $2$; this increases the total sum by $1$. We bring then the type I dice back to $6$, and repeat this sequence of moves until each type III die shows $4$, $5$ and $6$. We arrived thus to $S_{\max}$ and the proof is finished.