The term under summation is: $$f(n,k)=\sqrt[n]{\frac1{k^{n-1}}} =k^{\frac1n-1}$$
Note that: $$f(n,k-1)>f(n,k)$$
The total of the sum is the area under a decreasing stepwise function and is then bounded below and above by the integrals over the summation range of $f(n, k)$ and $f(n, k-1)$, respectively: $$\int_1^{p^n}k^{\frac1n-1}dk<\sum_1^{p^n}k^{\frac1n-1}<\int_1^{p^n}(k-1)^{\frac1n-1}dk$$
$f(n,k)$ passes through the left-hand points of the stepwise function, $f(n,k-1)$ passes through the right-hand points - so the summation is a left Riemann sum of $f(n,k)$ and a right Riemann sum of $f(n,k-1)$. The upper bound may be tightened by noting that the first term in the summation is $1$: $$\int_1^{p^n}k^{\frac1n-1}dk<\sum_1^{p^n}k^{\frac1n-1}<1 + \int_2^{p^n}(k-1)^{\frac1n-1}dk<1+\int_2^{p^n+1}(k-1)^{\frac1n-1}dk$$
Now evaluating the left and rightmost integrals:$$\int_1^{p^n}k^{\frac1n-1}dk=nk^\frac1n\big|_1^{p^n}=np-n=n(p-1)$$ $$1+\int_2^{p^n+1}(k-1)^{\frac1n-1}dk=1+\big(n(k-1)^\frac1n\big|_2^{p^n+1}\big)=1+np-n=1+n(p-1)$$
Therefore: $$n(p-1)<\sum_1^{p^n}k^{\frac1n-1}<1+n(p-1)$$
Since $n$ and $p$ are integers and $1+n(p-1)-n(p-1)=1$ we have that the lower bound is the integral part of the summation as required.
Solutions for the integral part being $2016$ then satisfy:$$n(p-1)=2016\space\forall n\in\Bbb N,p\space\text prime$$.
Yielding $17$ "relevant" primes (listed here with their respective $n$):
\begin{align}
p&=1009;&n&=2\\
p&=673;&n&=3\\
p&=337;&n&=6\\
p&=127;&n&=16\\
p&=113;&n&=18\\
p&=97;&n&=21\\
p&=73;&n&=28\\
p&=43;&n&=48\\
p&=37;&n&=56\\
p&=29;&n&=72\\
p&=19;&n&=112\\
p&=17;&n&=126\\
p&=13;&n&=168\\
p&=7;&n&=336\\
p&=5;&n&=504\\
p&=3;&n&=1008\\
p&=2;&n&=2016\\
\end{align}
Other relevant material related to this puzzle for those interested can be found on Wikipedia:
Harmonic series and Generalized harmonic numbers
Newton's identities
Euler-Mascheroni constant (the fractional part tends to this)
Credit to humn and ffao for elucidating my thoughts while I slept!
