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Let me start by rephrasing what is already in the answers of 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.

Side remark: Observe that the story for $E_0$ is even more degerated. Nevertheless,deformation theory of $E_0$-algebras is still very interesting (for a discussion about this issue and its relation to the BV formalisms, see Costello-Gwilliam work-in-progress http://math.northwestern.edu/~costello/factorization_public.html - especially 5b and 5c).

Let me start by rephrasing what is already in the answers of 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 invoke 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.

Side remark: Observe that the story for $E_0$ is even more degerated. Nevertheless,deformation theory of $E_0$-algebras is still very interesting (for a discussion about this issue and its relation to the BV formalisms, see Costello-Gwilliam work-in-progress http://math.northwestern.edu/~costello/factorization_public.html - especially 5b and 5c).

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.

Let me start by rephrasing what is already in the answers of 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.

Side remark: Observe that the story for $E_0$ is even more degerated. Nevertheless,deformation theory of $E_0$-algebras is still very interesting (for a discussion about this issue and its relation to the BV formalisms, see Costello-Gwilliam work-in-progress http://math.northwestern.edu/~costello/factorization_public.html - especially 5b and 5c).

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 $A=C_{-\bullet}(\Omega^dX,\mathbb{Q}$ equipped with the degree filtration. Then` $gr(A)=H_{-\bullet}(\Omega^dX,\mathbb{Q}$ is going to be a ${\rm pois}^d$-algebra.

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.