Let $f(x): \mathbf{C} \to \mathbf{C}$ be a complex valued polynomial of degree $n$, and suppose that for a random point $z$ sampled uniformly from the unit disk $|z|\leq 1$ we have
$$\mathbf{P}\_{z}( |f(z)| \leq \epsilon ) \geq 1 - n^{-k},$$
for some large positive $k>0$.
Under what conditions can we conclude that $|f(x)|\leq \epsilon$ for all $|x|\leq 1$?



Bounding function by random sampling
As pointed out by fedja, one can not expect the inequality yo hold with the same epsilon. However, the following holds : if $\mathbf{P}\_{z}( |f(z)| \leq \epsilon ) \geq 1 - \delta$, with $\delta \ll n^{-2} (\log n)^{-2}$, then
$$
\forall z \in \mathbb D, \ |f(z)| \leq \epsilon \left(1 + O\left(n \sqrt{\delta} \log \frac{1}{\delta}\right) \right)
$$
holds. In the other direction, Fedja's comment yields for each $\delta < 1$ an example where the sup norm is $\geq \epsilon( 1 + \frac{1}{2} n \delta)$.


**Proof**: assume we have proved an equality of the form $|| f||\_{L^{\infty}} \leq C\_{n,p} || f||\_{L^{p}}$ for some $p$. Then, writing
$$
C\_{n,p}^{-p}|| f||\_{L^{\infty}}^p \leq || f||\_{L^{p}}^p \leq \epsilon^p + \delta || f||\_{L^{\infty}}^p,
$$
and assuming $\delta C\_{n,p}^p < 1$, one gets
$$
|| f||\_{L^{\infty}} \leq \epsilon \frac{C\_{n,p}}{\left( 1 - \delta C\_{n,p}^p \right)^{\frac{1}{p}}}.
$$
Thus, one just needs good estimates on $C\_{n,p}$. Let us start with $p=2$. If $f(z) = \sum\_{i=0}^n a\_i z^i$, then
$$
|| f||\_{L^{\infty}} \leq \sum\_{i=0}^n |a\_i| \ \ \text{and} \ \ || f||\_{L^{p}}^2 = \sum\_{i=0}^n \frac{|a\_i|^2}{i+1},
$$
so that
$$
C\_{n,2}^2 = \sum\_{i=0}^n (i+1) = \frac{1}{2} (n+1) (n+2) \asymp n^2.
$$
Now, if $p = 2k$ for some integer $k$, applying the previous inequality to the polynomial $f^k$ shows that one can choose
$$
C\_{n,2k}^{2k} = C\_{nk,2}^2 \asymp n^2 k^2.
$$ 
One then takes $k \asymp n^{-1} \delta^{-\frac{1}{2}}$ so that $\delta C\_{n,p}^p \leq \frac{1}{2}$ for $p=2k$. We then have
$$
C\_{n,p} = 1 + O\left(n \sqrt{\delta} \log \frac{1}{\delta}\right),\\
\left( 1 - \delta C\_{n,p}^p \right)^{\frac{1}{p}} = 1 + O\left(n \sqrt{\delta} \right),
$$
hence the result.


