Yes, such $c_n$ is bounded by something effective. Below is a cubic bound, which probably may be improved.

Assume that $f(x)$ is a power of 2 for all integer $x$ on $[0,c_n]$.
Note that $[0,c_n]$ is partitioned onto at most $n$ segments, onto each of which $f$ is monotone. Thus there exist $N:=(c_n+1)/n$ consecutive integers onto which the values of $f(x)$ are, say, increasing powers of 2: $2^{m_1}<2^{m_2}<\ldots<2^{m_N}$. Assume that $N>(n+1)^2$. Denote $p_j=m_{1+j(n+1)}$ for $j=0,1,\ldots,n+1$. The numbers $2^{p_j}$ are the values of a polynomial of degree $n$ along $n+2$ elements of an arithmetic progression. Thus $$2^{p_{n+1}}-{n+1\choose 1}2^{p_n}+{n+1\choose 2}2^{p_{n-1}}-\ldots=0.$$
But the first summand is greater than the sum of all others.