I'd like to attempt an answer. I will have no problem with down-votes or with deleting it, since I am not 100% confident in it.
Let $A$ be the annulus $\frac{2}{3} \le |z| \le \frac{3}{2}$. For a sequence $a_n \in \mathbb{C}$, define $f_{a_n}(z) = f(a_n z)$ and $B_{a_n} = \{a_n z : z \in A \}$.
$B_{a_n}$ is an annulus with inner radius $|a_n|\frac{2}{3}$ and outer radius $|a_n|\frac{3}{2}$.
Let $a_n = n$, n positive integer. Let $f_{n_k}$ be the subsequence of $f_n$ guaranteed by the statement. Then either $f_{n_k} \to \infty$ uniformly on $A$ or $f_{n_k} \to g$ uniformly on $A$ for an analytic function $g$.
In the second case, $g$ is analytic so $g(A)$ is bounded. Using uniform convergence, you can see that There is an $M$ such that $f_{n_k}(A) < M$ for all $k$ sufficiently large. This means $f(z) < M$ for all $z \in B_{n_k}$, with $k$ sufficiently large. Mimicking the proof of Liouville's theorem that uses the Cauchy Integral Formula, this shows that $f(z)$ is constant.
In the first case, for any $M$ you can achieve $|f_{n_k}(z)| > M$ for $k$ sufficiently large. If in addition, for $k$ sufficiently large, $B_{n_k} \cap B_{n_{k+1}}$ is non-emtpy, then this shows $f(z) \to \infty$ as $z \to \infty$, and therefore $f$ must be a polynomial.
If on the other hand there are infinitely many $k$ with $B_{n_k} \cap B_{n_{k+1}}$ empty, then we need to look at what happens in these "gaps". Note that the outer radius of $B_{n_k}$ is in this case strictly less than the inner radius of $B_{n_{k+1}}$. This "gap" is an open annulus, but $f$ is analytic on the closure. Suppose such a gap does not contain a zero of $f$. Then the Minimum Modulus Theorem says that the modulus of $f$ in the gap is at least as great as the minimum modulus on the boundary, and so in the gap $f|(z)| > M$.
Therefore if only finitely many gaps contain zeros, we again see that $f(z) \to \infty$ as $z \to \infty$.
Finally, suppose there are infinitely $n_k$ with $B_{n_k} \cap B_{n_{k+1}}$ empty, and infinitely many of these "gaps" contain a zero of $f$. For the $m$-th gap pick a zero $b_m$, with $|b_m| < |b_{m+1}|$ and $|b_m| \to \infty$. These $b_m$ define a new sequence $f_{b_m}$ of functions on $A$, a subsequence of which converges uniformly. But in this case it cannot converge uniformly to $\infty$, because each function in the sequence has a zero at $z = 1$. Then as in the proof of the second case, $f$ must be constant.
Best Answer
If $f$ is not identically zero then $f$ has a zero of multiplicity $k \ge 0$ at $z=0$, i.e. $$ f(z) = z^k g(z) $$ where $g$ is an entire function with $g(0) \ne 0$.
Then $$ \bigl|g(\frac 1n)\bigr| = n^k \, \bigl|f(\frac 1n)\bigr| \le \frac{n^k}{n^n} $$ for all positive integers $n$. The right-hand side tends to zero for $n \to \infty$ and it follows that $g(0) = 0$, which is a contradiction.