There is a projective resolution $P_\bullet$ of $\mathbb Z$ as an $F_n$-module of the form $$0\to\mathbb ZF_n\otimes V\to\mathbb ZF_n\to\mathbb Z\to0$$ in which $V$ is the free $\mathbb Z$-module spanned by the generators $x_1,\dots,x_n$ of $F_n$, and the differential is the $\mathbb ZF_n$-linear map such that $d(1\otimes x_i)=x_1-1$.

On the other hand, you have the bar resolution $B_\bullet$ of $\mathbb Z$ as a $\mathbb ZF_n$-module. The two complexes are homotopic, so that there are maps $\phi:P_\bullet\to B_\bullet$ and $\psi:B_\bullet\to P_\bullet$ such that $\phi\circ\psi$ and $\psi\circ\phi$ are homotopic to identity maps. If you make explicit an homotopy $s:\phi\circ\psi\simeq 1_{B_\bullet}$, you can do what you want: if $\alpha$ is a $p$-cocycle on the bar complex and $p>1$, then the boundary of $\alpha\circ s$ is $\alpha$.

There is a projective resolution $P_\bullet$ of $\mathbb Z$ as an $F_n$-module of the form $$0\to\mathbb ZF_n\otimes V\to\mathbb ZF_n\to\mathbb Z\to0$$ in which $V$ is the free $\mathbb Z$-module spanned by the generators $x_1,\dots,x_n$ of $F_n$, and the differential is the $\mathbb ZF_n$-linear map such that $d(1\otimes x_i)=x_i-1$.

On the other hand, you have the bar resolution $B_\bullet$ of $\mathbb Z$ as a $\mathbb ZF_n$-module. The two complexes are homotopic, so that there are maps $\phi:P_\bullet\to B_\bullet$ and $\psi:B_\bullet\to P_\bullet$ such that $\phi\circ\psi$ and $\psi\circ\phi$ are homotopic to identity maps. If you make explicit an homotopy $s:\phi\circ\psi\simeq 1_{B_\bullet}$, you can do what you want: if $\alpha$ is a $p$-cocycle on the bar complex and $p>1$, then the boundary of $\alpha\circ s$ is $\alpha$.