Let $\mu_n$ be the standard normal distribution over $\mathbb{R}^n$. Is the following true?
For all $\epsilon > 0$ and $d \in \mathbb{N}$, there exists $\eta > 0$ such that for all $n \geq 1$ and $T \in \mathcal{B}_{\mathbb{R}^n}$ with $\mu_n(T) \leq \eta$, the following holds. For all polynomials $f:\mathbb{R}^n \to \mathbb{R}$ of degree at most $d$, $\int_{T} |f|d\mu_n \leq \epsilon \int |f| d\mu_n$
If so, what is a sufficient condition on the class of distributions $\{\mu_n\}_{n \geq 1}$ for this statement to hold?
In this similar question, $\eta$ is allowed to depend on $f$.
Here is a solution for fixed $n$.
Fix $d$. What you want is equivalent to the following: for every $\epsilon >0$, there exists $\eta >0$ such that for every $T \in \mathcal{B}(\mathbb{R}^n)$ with $\mu_n(T)\leq \eta$, we have $\sup_{f\in \Phi} \int_T |f|d\mu_n\leq \epsilon$ where I set $$\Phi = \left\{ f \in \mathbb{R}_d[X],\,\int|f| d\mu_n\leq 1\right\}.$$ This property is sometimes called "uniform continuity" of the family $\Phi$. Since $\Phi$ is bounded in $L^1(\mu_n)$, this is equivalent to the uniform integrability of $\Phi$. By the Dunford-Pettis theorem, this is equivalent to the family $\Phi$ being relatively compact for the weak topology of $L^1(\mu_n)$. Now observe that $\Phi$ is the closed unit ball of the finite-dimensional space $\mathbb{R}_d[X]$ (equipped with the $L^1(\mu_n)$ norm). As such, $\Phi$ is in fact strongly compact for said norm and in particular it is weakly compact. This proves the result.
This proof is non constructive and possibly uses results which are too strong. Notice however that we did not need any particular properties related to the Gaussian measure $\mu_n$ so it should work for any probability measure $\mu$ with respect to which all polynomials are integrable.