Second homology group of free nilpotent p-group Let $F_n$ be a free group on $n$ generators.  Fix a prime $p$.  Let $\gamma_k^p(F_n)$ be the mod $p$ lower central series, i.e. the inductively defined series
$$\gamma_0^p(F_n) = F_n \quad \text{and} \quad \gamma_{k+1}^p(F_n) = (\gamma_{k}^p(F_n))^p [F_n,\gamma_k^p(F_n)].$$
Observe that the quotients $\gamma_{k}^p(F_n) / \gamma_{k+1}^p(F_n)$ are abelian $p$-groups.  Moreover, the quotients $N_n^p := F_n / \gamma_{k}^p(F_n)$ are $p$-groups of nilpotency class $k$.  They are universal with this property -- if $G$ is a $p$-group of nilpotency class $k$ and $g_1,\ldots,g_n \in G$, then there is a unique homomorphism $N_n^p \rightarrow G$ taking the generators of $N_n^p$ to the $g_i$.
Question : What are $H_2(N^p_n;\mathbb{Z})$ and $H_2(N^p_n;\mathbb{F}_p)$?
 A: 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.    
