jmc

Let the values on one pair of opposite faces be $a$ and $d$; the second pair of faces, $b$ and $e$, and the third pair of faces, $c$ and $f$. There are eight vertices on the cube, so we find that the sum 1001 is equal to $$abc+aec+abf+aef+dbc+dec+dbf+def.$$ For any two faces adjacent at a vertex with $a$, the same two faces are adjacent to a vertex with $d$. Furthermore, any three adjacent faces must contain one of $a$ or $d$. Therefore, every term contains $a$ or $d$, and the expression is symmetric in $a$ and $d$. Considering the expression as a polynomial in $a$ (with the remaining variables fixed), we observe that $P(-d)=0$. Therefore, $a+d$ divides the given expression. Similarly, $b+e$ and $c+f$ divide the given expression as well. Therefore, $$abc+aec+abf+aef+dbc+dec+dbf+def=k(a+d)(b+e)(c+f).$$ Here, since both sides are of degree three in their variables, $k$ must be a constant, which is easily seen to be $1$.

It follows that $(a+d)(b+e)(c+f)=1001=7\cdot 11\cdot 13$. Since each of the variables is positive, we have $a+d>1,b+e>1,$ and $c+f>1$. Thus $(a+d)+(b+e)+(c+f)=7+11+13=\boxed{31}$.