For brevity, I denote by $F$ the free group on $n$ generators, $\lambda_k$ the $k$th terme of the $p$-lower series of $F$, and $N_k$ the relatively free group $F/\lambda_k$.

Also I use the following 5- terme exact sequence (see Generators of sections of free groups): $$0 \longrightarrow H_2(N_k;\mathbb{Z}/p) \longrightarrow \lambda_k/[F,\lambda_k]{\lambda_k}^p \longrightarrow H_1(F;\mathbb{Z}/p) \longrightarrow H_1(N_k;\mathbb{Z}/p) \longrightarrow 0.$$ Now since $N_k$ is minimally $n$-generated $p$-group, it follows from $ H_1(N_k;\mathbb{Z}/p) = N_k/[N_k,N_k]{N_k}^p$ that $ H_1(N_k;\mathbb{Z}/p) \cong (\mathbb{Z}/p)^n$, and also we have $H_1(F;\mathbb{Z}/p) = F/[F,F]F^p \cong (\mathbb{Z}/p)^n$. Now the above exact sequence reduces to $$0 \longrightarrow H_2(N_k;\mathbb{Z}/p) \longrightarrow \lambda_k/[F,\lambda_k]{\lambda_k}^p \longrightarrow 0$$

so we have $H_2(N_k;\mathbb{Z}/p) \cong \lambda_k/\lambda_{k+1}$, and this is completely determined by the rank $r_k$ of the elementary abalian $p$-group $\lambda_k/\lambda_{k+1}$.

The integer $r_k$ can be determined using Lie methods, see for instance Corollary 18 in "The automorphism group of a finite p-group is almost always a p-group" Journal of Algebra, Volume 312 (2007) G. T. Helleloid and U. Martin. In which it is proved that $$r_k = \sum_{i=1}^{k} 1/i (\sum_{j/i} \mu(i/j) n^j)$$ where $\mu(t)$ denotes the Mobius function.

For brevity, I denote by $F$ the free group on $n$ generators, $\lambda_k$ the $k$th terme of the $p$-lower series of $F$, and $N_k$ the relatively free group $F/\lambda_k$.

Also I use the following 5- terme exact sequence (see Generators of sections of free groups): $$0 \longrightarrow H_2(N_k;\mathbb{Z}/p) \longrightarrow \lambda_k/[F,\lambda_k]{\lambda_k}^p \longrightarrow H_1(F;\mathbb{Z}/p) \longrightarrow H_1(N_k;\mathbb{Z}/p) \longrightarrow 0.$$ Now since $N_k$ is minimally $n$-generated $p$-group, it follows from $ H_1(N_k;\mathbb{Z}/p) = N_k/[N_k,N_k]{N_k}^p$ that $ H_1(N_k;\mathbb{Z}/p) \cong (\mathbb{Z}/p)^n$, and also we have $H_1(F;\mathbb{Z}/p) = F/[F,F]F^p \cong (\mathbb{Z}/p)^n$. Now the above exact sequence reduces to $$0 \longrightarrow H_2(N_k;\mathbb{Z}/p) \longrightarrow \lambda_k/[F,\lambda_k]{\lambda_k}^p \longrightarrow 0$$

so we have $H_2(N_k;\mathbb{Z}/p) \cong \lambda_k/\lambda_{k+1}$, and this is completely determined by the rank $r_k$ of the elementary abalian $p$-group $\lambda_k/\lambda_{k+1}$.

The integer $r_k$ can be determined using Lie methods, see for instance Corollary 18 in "The automorphism group of a finite p-group is almost always a p-group" Journal of Algebra, Volume 312 (2007) G. T. Helleloid and U. Martin. In which it is proved that $$r_k = \sum_{i=1}^{k} 1/i (\sum_{j/i} \mu(i/j) n^j)$$ where $\mu(t)$ denotes the Mobius function.