Aren't there?

The expression factors into

$n^3+4n^2-20n+48=(n-4)(n+2)(n+6)$ .

For this to be divisible by $125=5^3$ , there have to be at least 3 factors of 5 in the product.

If $(n+2)$ is divisible by 5, then $(n-4)$ and $(n+6)$ aren't so all three 5's have to come from $(n+2)$ , which would make $n$ at least 125.

If $(n-4)$ is divisible by 5, then so is $(n+6)$ , while $(n+2)$ isn't. To have three factors of 5, one of the two has to be divisible by 25, but to minimize $n$ , this factor should be $(n+6)=25$ .

So, $n=\boxed{19}$ , and $125 | 15 \cdot 21 \cdot 25$ .