You could use dynamical programming. Let $a(m,n)$ be the number of $m$-digit nonnegative integers with each digit 0 to 7 and the sum of digits is $n$. Then $a(1,n) = 1$ for $0 \le n \le 7$, $0$ otherwise, and $a(m+1,n) = \sum_{d=0}^{\min(n,7)} a(m,n-d)$
EDIT: That solution allows leading zeros. If leading zeros are not allowed, take $a(1,0) = 0$ as well.
EDIT: Let $a(m,n)$ be as above in the version with leading zeros not allowed, and $b(m,n)$ the version with leading zeros allowed. Then $a(6,20) = \sum_{k=0}^{20} a(3,k) b(3,20-k)$. For $m = 1,2,3$ we make the following tables:
n a(m,n) b(m,n) m=1 2 3 1 2 3 0 0 0 0 1 1 1 1 1 1 1 1 2 3 2 1 2 3 1 3 6 3 1 3 6 1 4 10 4 1 4 10 1 5 15 5 1 5 15 1 6 21 6 1 6 21 1 7 28 7 1 7 28 1 8 36 8 0 7 35 0 7 42 9 0 6 40 0 6 46 10 0 5 43 0 5 48 11 0 4 44 0 4 48 12 0 3 43 0 3 46 13 0 2 40 0 2 42 14 0 1 35 0 1 36 15 0 0 28 0 0 28 16 0 0 21 0 0 21 17 0 0 15 0 0 15 18 0 0 10 0 0 10 19 0 0 6 0 0 6 20 0 0 3 0 0 3
Then (going down the $a(3,n)$ column and up the $b(3,n)$ column) your answer is $\eqalign{&0 \times 3 + 1 \times 6 + 3 \times 10 + 6 \times 15 + 10 \times 21 + 15 \times 28 + 21 \times 36 + \cr &28 \times 42 + 35 \times 46 + 40 \times 48 + 43 \times 48 + 44 \times 46 + 43 \times 42 + 40 \times 36 + \cr &35 \times 28 + 28 \times 21 + 21 \times 15 + 15 \times 10 + 10 \times 6 + 6 \times 3 + 3 \times 1 \cr}$