Here's a counterexample with $d=1$. Since $d=1$, we have $p_n(x)=a_nx^n$, so $p(x)=\sum_{n=0}^\infty a_nx^n$. Since you call this a polynomial, we'll also assume that $a_n=0$ for sufficiently large $n$. Let's take a special case of what you'd like to be true:
$$
|p_2(1)|\leq C\sup_{x\in K}|p(x)|
$$
for some compact $K\subseteq\mathbb R$ and some $C<\infty$. In other words:
$$
|a_2|\leq C\sup_{|x|\leq N}|p(x)|
$$
Suppose this is true for all polynomials $p(x)$, and apply it to a partial taylor expansion of $f_A(x)=e^{-Ax^2}$, $A>0$. Then $a_2=\frac 12f_A''(0)=-A$, so your inequality would imply that $|A|\leq C\sup_{|x|\leq N}|p(x)|$. However the power series of $f_A(x)$ has infinite radius of convergence, so no matter what $N$ is, we can choose a very high degree taylor approximation which equals $f_A$ within an error of $\frac 1{10}$ on $[-N,N]$. Thus the right hand side will be bounded by, say $2C$. On the other hand, we are free to choose $A$ as large as we want, so this is a contradiction.

There are two ways I see to fix the desired inequality. First, you could assume bound the degree of $p(x)$. Then the space of such polynomials is finite dimensional, and in this case any $K$ with nonempty interior works, since then $\sup_{z\in K}|p(z)|$ is a norm on this space of such polynomials. (incidentally, this is essentially related to Problem A5 on the 1999 Putnam, see http://amc.maa.org/a-activities/a7-problems/putnamindex.shtml).

Second, you could instead consider polynomials on the complex numbers, and allow supremum on the right hand side to be over compact $K\subseteq\mathbb C^n$. Then the Cauchy Integral Formula (http://en.wikipedia.org/wiki/Cauchy_integral_formula) would make your statement true (at least for $d=1$, and probably all $d$ as well).

`$p_n$`

while the right side involves $p$. $\endgroup$ – Andreas Blass May 18 '11 at 17:45