Comparing the truncated $\ell^{1}$-norm of polynomial coefficients with the supremum norm on the unit disc. Let $p=a_{0}+a_{1}z+\ldots+a_{n}z^{n}$ be a polynomial. Consider the following truncated $\ell^{1}$-seminorm of the coefficients of $p$:
$$\|p\|_{\ell^{1},\text{trun.}}:=\sum_{k=1}^{n}|a_{k}|=\|p-a_{0}\|_{\ell^{1}}.$$
Does there exist $C>0$ (independent of $n=\deg(p)$) such that
$$\|p\|_{\ell^{1},\text{trun.}}\leq C\|p\|_{\infty}\qquad\text{where}\qquad \|p\|_{\infty}:=\sup_{|z|\leq1}|p(z)|?\tag{$*$}$$
Remark 1: $\sup_{|z|\leq1}|p(z)|=\sup_{|z|=1}|p(z)|$ by the maximum modulus principle.
Remark 2: If $p$ is a polynomial with $a_{j}\geq0$, then the inequality clearly holds with $C=1$. Indeed, $\|p\|_{\ell^{1},\text{trun.}}\leq p(1)\leq\|p\|_{\infty}$.
Remark 3: Due to this post I believe that estimation ($*$) does $C$ does not exist independent of $n$ if $\|\cdot\|_{\ell^{1},\text{trun.}}$ is replaced by $\|\cdot\|_{\ell^{1}}$.
Remark 4: The same MSE post also provides an example that shows $C>1$ if it exists. Namely, if $p(z)=z^{2}+2iz+1$, then $\|p\|_{\ell^{1},\text{trun.}}=|1|+|2i|=3$ and $\|p\|_{\infty}=2\sqrt{2}$ (I haven't checked this last fact).
 A: The answer to the question is negative.
Assume for a contradiction that
$$\|p-a_0\|_{\ell^1}\le C\|p\|_\infty$$
As $|a_o|=|p(0)|\le \|p\|_\infty$ we get
$$\|p\|_{\ell^1}\le D\|p\|_\infty,\qquad D=C+1$$
By the maximum principle we have
$$\|p\|_\infty=\max_{|z|=1|}|p(z)|=\max_{0\le t<2\pi}|p(e^{it})|$$
Therefore for any trigonemtric polynomial
$q(t)=\displaystyle\sum_{k=-n}^na_ke^{ikt}$ we have
$$\sum_{k=-n}^n |a_k| =\|e^{int}q(t)\|_{\ell^1}\le D\|e^{int}q(t)\|=D\|q(t)\|_\infty \qquad (*) $$
By the Weierstrass theorem trigonometric polynomials are uniformly dense in the space $C_{per}(\mathbb{R})$ of continuous periodic functions with period $2\pi.$  The inequality $(*)$ would imply that any continuous periodic function has absolutely convergent Fourier series. This gives a contradiction.
More explicit explanation: the trigonometric polynomials
$$\sum_{k=1}^n{\sin kx\over k}=\sum_{k=1}^n{e^{ikx}-e^{-ikx}\over 2ki}$$  are bounded independently of $n$ but the $\ell^1$ norms are proportional to $\log n.$
