Letting all $a_i=0$, we find $$f(0)=\frac nk f(0)^n$$hence either $f(0)=0$ or $f(0)=\sqrt[2013]{2014^{2012}}$, but the latter is not $\in\mathbb Z^+$. We conclude $f(0)=0$.
Letting all $a_i=0$ except $a_1=1$, we find
$$ f(1)= \frac1kf(1)^n$$
hence either $f(1)=0$ or $f(1)=2014$.
In the latter case, we find $f(2)=\frac{2014^{2014}+2014^{2014}}{2014^{2013}}=2\cdot 2014$ and in fact $f(a)=2014a$ for $0\le a\le n$.
In the first case we immediately find $f(2)=f(3)=\ldots = f(2014)=0$.
In fact, we immediately check that the two functions $f(a)=0$ and $f(a)=2014 a$ are valid solutions (in other words, $f(a)=af(1)$ and $f(1)$ is one of the two possible values found above).
We suspect that there is no third solution and observe that we did not make use of the growth condition yet.
Let $$S=\{\,a\in \mathbb Z^+\mid f(a)=af(1)\,\}$$Then $0,1,\ldots, 2014\in S$ as seen above. Moreover, if $a_1,\ldots, a_n\in S$ then $\sum a_i^n\in S$, hence $S$ is not bounded from above.
If $f(1)=0$ this gives us immediately that $f(a)=0$ for all $a$ because $f$ is nondecreasing.
Thus we may assume from now on that $f(1)=2014$.
For $a\in\mathbb N$ let $q(a) =\frac{f(a)}{2014a}$.
From $f(ra^n)=\frac1k{rf(a)^n}=\frac{rq(a)^na^n2014^n}{2014^{2013}}=2014q^nra^n$ we see $q(ra^n)=q(a)^n$ for $1\le r\le 2014$. Hence, unless $f$ is linear, these $q(a)$ become arbitrarily big or arbitrarily small. Show that the set $S$ is "sufficiently" dense to disallow very small/big values of $q(a)$.
