Do not add every number individually.

For each of the $9$ "hundreds" digits, there are $9$ non-zero "tens" digits, and for each of these there are $9$ non-zero "ones" digits. Since $\sum_{n=1}^9 n = 45$ , the sum of all the products of the digits will then be $45^{3}$ . Calculating the desired remainder, we observe that

$45^{3} \equiv (-4)^{3} \equiv -64 \equiv 6 \mod{7}$ .

The desired remainder is thus $\boxed{6}$ .