Let me start by rephrasing what is already in the answers David Ben-Zvi and Theo Johnson-Freyd. The DG $\mathbb{Q}$-linear operad $\mathbb{E}_n:=C_{-\bullet}(E_n,\mathbb{Q})$ is filtered. For $n\geq2$ the filtration is the degree filtration, and thus $gr(\mathbb{E}_n)=H_{-\bullet}(E_n,\mathbb{Q})={\rm Pois}^n$. 

The situation for $n=1$ is a bit different. We know that $\mathbb{E}_1\cong {\rm As}$ (this is the formality theorem for $E_1$ which, contrary to the case when $n\geq2$, is easy to prove). The operad ${\rm As}$ of associative algebras is also filtered, but in a less obvious way. To be short, one assigns the following  two-step filtration onto ${\rm As}(2)=\mathbb{Q}[\Sigma_2]$ (which generates ${\rm As}$): 
$$
F^0{\rm As}(2)=\mathbb{Q}(1-\sigma)\subset F^1{\rm As}(2)={\rm As}(2).
$$
It then an exercise to check that $gr({\rm As})={\rm Pois}^1$. 

Then, in order to relate the two stories, I have the feeling that one does not need to invoque the formality of $E_n$ for $n\geq2$. Given a filtered $\mathbb{E}_n$-algebra $A$ (i.e. a filtered DG $\mathbb{Q}$-vector space equipped with an action of $\mathbb{E}_n$ that is compatible with the above filtration), then $gr(A)$ is a ${\rm Pois}^n$-algebra. 

Concerning the last example in the question, one has to take $A=C_{-\bullet}(\Omega^d(X),\mathbb{Q})$ equipped with the degree filtration. Then $gr(A)=H_{-\bullet}(\Omega^d(X),\mathbb{Q})$ is going to be a ${\rm Pois}^d$-algebra.