2016

$2016^n+\underbrace{2016^2+2016^2+\ldots+2016^2}_{p\text{ times}}=2016^3\\ 2016^n+p(2016^2)=2016^3\\ p(2016^2)=2016^3-2016^n\\ p=\dfrac{2016^3-2016^n}{2016^2}\\ p=2016-2016^{n-2}$

Now, given that $p$ is positive, we know that

$2016-2016^{n-2}>0\\ 2016^1>2016^{n-2}\\ 1>n-2\\ n<3$

Given that $p$ is an integer, we know that $2016^{n-2}$ must be an integer. This implies that:

$n-2 \geq 0\\ n \geq 2$

Combine the two conditions, we have

$2 \leq n < 3$

For this range of values of $n$ , there is only one integer value that $n$ can take, and that is $n=\boxed{2}$