My question is quite simple:
Is there a constant $C>0$ such that for all complex polynomials $p(z)=\sum_{n=0}^N a_n z^n, a_n \in \mathbb{C}$ $$ \sum_{n=0}^N|a_n|\leq C \sup_{z\in \overline{\mathbb{D}}} |p(z)|$$ holds? I have found counterexamples for e.g. $C=1$, but I am stuck at the general case.
Consider for given $N$, $d_N>0$ the minimum of all constants s.t. $\sum_{n=0}^N|a_n|\leq d_N\sup_{z\in \overline{\mathbb{D}}} |p(z)|=||p||$ for any polynomial $p(z)=\sum_{n=0}^N a_n z^n, a_n \in \mathbb{C}$ of degree at most $N$. Since by scaling we can reduce the problem to polynomials for which $||p||=1$, we can also define $d_N =\max_{||p||=1, \deg p \le N}\sum_{n=0}^N|a_n|$.
It is trivial that $d_N \le N+1$, by for example Cauchy formula since $|a_k| \le \sup_{z\in \overline{\mathbb{D}}} |p(z)|, k=0,...N$ and it also easy to show that $d_N \le \sqrt{N+1}$ using Parseval and Cauchy-Schwarz since:
$\sum{|a_k|^2}=||p||_2^2 \le ||p||^2$ and obviously $(\sum{|a_k|})^2 \le (N+1)\sum{|a_k|^2}$, where as usual $||p||_2^2=\frac{1}{2\pi}\int_0^{2\pi}|p(e^{it})|^2dt$
Using slightly more advanced techniques like Visser inequality $|a_0|+|a_N| \le ||p||$ and Cauchy-Schwarz again we get $d_N \le \sqrt N$ and it is easy to see (checking the equality case in Cauchy Schwarz) that we have strict inequality for $N \ge 5$ (and $N=3$, but we have equality for $N=1,2,4$).
Conversely it was proven by Beller and Newman (proc AMS 1971) that for $N$ large enough $d_N \ge \sqrt{N}-3N^{\frac{3}{10}}\log N$, so indeed $d_N$ is asymptotically $\sqrt N$
(edit later - for example in degree 2 it is easy to see that $p(z)=1+2iz+z^2$ gives equality for $d_2=\sqrt 2$ since $||p||=2\sqrt 2$, while in higher degrees one should consult the paper of Beller and Newman for examples of such polynomials for which $d_N$ gets very close to $\sqrt N$)