When $p = 2$, $n = 63$. When $p = 3$, $n = 103$. When $p = 5$, $n = 567$.

$$\begin{aligned} (p^2 + 2)^2 - 9(p^2 - 7) &= p^4 - 5p^2 + 4 + 63 \\ &= (p^2 - 1)(p^2 - 4) + 63 \\ &= (p - 1)(p + 1)(p - 2)(p + 2) + 63 \\ &= (p - 2)(p - 1)(p + 1)(p + 2) + 63 \end{aligned}$$

When $p \neq 3$, $(p-2)(p-1)$ and $(p+1)(p+2)$ are both multiples of 3. Thus $n = (p^2 + 2)^2 - 9(p^2 - 7)$ is a multiple of 9 and so the digit sum is a multiple of 9. When $n = 3$, the digit sum is 4. Thus this is the smallest value that can be obtained.