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Sure, there is. In fact given any finitely number of points on the boundary, say $z_1,\ldots,z_n$, there exists a polynomial such that $|p(0)|<|p(z_k)|$ that is zero free on the unit disc. An observation is that it follows by looking at the meromorphic function $$ f_{\epsilon}(z)=\prod_{k=1}^n z_k(z-z_k(1+\epsilon))^{-1} $$ By the construction $\lim_{\epsilon \to 0^+} f_\epsilon(0)=1$ and $\lim_{\epsilon \to 0^+} |f_\epsilon(z_k)|=\infty$. Thus we can choose an $\epsilon>0$ such that $|f_\epsilon(0)|<3/2<2<|f_\epsilon(z_k)|$. Since $f_\epsilon(z)$ is continuous and zero free on the closed unit square and analytic in the open unit square, a variant of Mergelyan's theorem of mine, http://arxiv.org/abs/1010.0850 shows that we can approximate the function arbitrarily closely (in sup norm) on the unit disc by a polynomial without zeros (this is where my variant is needed) in the unit square. If we find such a polynomial $p(z)$ that approximates the function $f_\epsilon(z)$ with an error less that $1/4$ then the inequality $|p(0)|<|p(z_k)|$ holds.