Let the numbers on the pair of opposite faces be $a_1, a_2, b_1, b_2, c_1, c_2$. Note that $a_1$ participates in 4 products: $a_1b_1c_1, a_1b_1c_2, a_1b_2c_1, a_1b_2c_2$. Likewise $a_2$ participates in the remaining 4 products and in a completely analogous fashion the products are $a_2b_1c_1, a_2b_1c_2, a_2b_2c_1, a_2b_2c_2$. It follows that the 8 products have sum

$$a_1(b_1c_1 + b_1c_2 + b_2c_1 + b_2c_2) + a_2(b_1c_1 + b_1c_2 + b_2c_1 + b_2c_2) = (a_1 + a_2)(b_1 + b_2)(c_1 + c_2).$$

Then by hypothesis $(a_1 + a_2)(b_1 + b_2)(c_1 + c_2) = 315$. Note that each factor on the left is greater than 1.
Thus $a_1 + a_2, b_1 + b_2, c_1 + c_2$ are divisors of 315 that are greater than 1 and have product 315.
Conversely, let $d_1, d_2, d_3$ be divisors of 315 with these properties. Since $d_i > 1$, one can write
$d_1 = a_1 + a_2, d_2 = b_1 + b_2, d_3 = c_1 + c_2$ with $a_1, a_2, b_1, b_2, c_1, c_2$ natural numbers. Place them on the faces of the cube (so that the $a_i$'s are on opposite faces; the same for the $b_i$'s and the $c_i$'s). Then, by the above, the sum of the 8 vertex products $a_i b_j c_k$ equals $(a_1 + a_2)(b_1 + b_2)(c_1 + c_2) = d_1 d_2 d_3 = 315$.

Note that the representations $d_1 = a_1 + a_2, d_2 = b_1 + b_2, d_3 = c_1 + c_2$ can be chosen in different ways. But the sum $a_1 + a_2 + b_1 + b_2 + c_1 + c_2$ which we are interested in is the same, equal to $d_1 + d_2 + d_3$ and depending only on the triple of divisors $d_1, d_2, d_3$ with the properties stated above.
In this way the question reduces to finding all factorizations $315 = d_1d_2d_3$ with factors $d_i > 1$. All solutions to the problem are the respective sums $D = d_1 + d_2 + d_3$. Since $315 = 3^2 \cdot 5 \cdot 7$, consider two cases. If one of the $d_i$'s be divisible by $3^2 = 9$ then it is immediate that the factorization is $5 \cdot 7 \cdot 9$ and $D = 5 + 7 + 9 = 21$. Otherwise two $d_i$'s are exactly divisible by 3, and it is clear that one of them is in fact equal to 3. It remains to factorize $3 \cdot 5 \cdot 7$ into two factors greater than 1, which can be done in 3 ways: $15 \cdot 7 \cdot 3$, $35 \cdot 3 \cdot 5$, $21$. Hence the remaining possibilities for admissible factorizations of 315 are $3 \cdot 15 \cdot 7$, $3 \cdot 35 \cdot 3$, $5 \cdot 21$. They yield respectively $D = 25, 41, 29$. In summary the answer to the problem is 21, 25, 29, and 41.