Edit: I had the feeling my original answer below was much too complicated. Here is a better answer.
New answer: Given $n$ complex numbers $z_k$ and $a_k$, $k=1,\ldots,n$ we can use standard interpolation arguments to find a polynomial $q(z)$ of degree $n$ such that $q(z_k)=a_k$. For example by choosing $a_k=k$ we will get $|q(z_k)|<|q(z_j)|$ iff $ k < j$. For any bounded set $D$, we can now add a large positive constant $K$ such that $p(z)=q(z)+K$ is zero free on $D$. In your example we can choose $z_1=0$ and $z_2,\ldots,z_9$ the points on the boundary of the unit square, and we can choose the constant $K$ sufficiently large so that $p(z)$ is zero free on the unit square and we get that $|p(0)|<|p(z_k)|$ for these points
Old answer: 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 square 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.