If $c_0=0$, $f(0)=c_n$ is qualified. Now assume $c_0\ne0$. Let $\alpha=0$ when $c_n=0$ whereas $\frac{c_n}{c_0}=re^{i\alpha}$ when $c_n\ne0$. Let $z_i,i=1,\cdots,n$ be the $n$ roots to $z^n=1$. $|\sum_{i=1}^n f(z_ie^{i\frac{\alpha}n})|=|nc_0e^{i\alpha}+nc_n|=n(|c_0|+|c_n|)\le\sum_{i=1}^n |f(z_ie^{i\frac{\alpha}n})|.$ So there exists $|f(z_ie^{i\frac{\alpha}n})|\ge |c_0|+|c_n|$ for some $i$. Also $|z_ie^{i\frac{\alpha}n}|=1$.

I really like this problem as a 1/4; it's fairly easy but well-motivated and satisfying to solve. To make sense of the $|c_0|$ and $|c_n|$ terms, we use roots of unity filter. Actually, let us fix a complex number $\alpha$ with $|\alpha| = 1$, and write $$\sum_{\omega^n = 1} f(\alpha \omega) = n(c_n \alpha^n + c_0)$$by roots of unity filter. To split the right side into a sum of absolute values, the purpose of $\alpha$ now becomes clear: let $n \arg \alpha + \arg c_n = \arg c_0$. Then $$\sum_{\omega^n = 1} |f(\alpha \omega)| \geq \left|\sum_{\omega^n = 1} f(\alpha\omega)\right| = |n(c_n\alpha^n + c_0)| = n(|c_n \alpha^n| + |c_0|) = n(|c_0| + |c_n|),$$so by Pigeonhole, there exists one of these terms $$|f(\alpha \omega)| \geq |c_0|+|c_n|,$$as desired.