Suppose that $\Gamma_n$ is the free nilpotent group on $k$ generators of nilpotency class $n$. Any finite $k$-generated $p$-group ($p$ a prime) is nilpotent, hence the free $k$-generated pro-$p$ group is surjected by the (universal) inverse limit of $\Gamma_n$. Hence any inverse limit of $\Gamma_n$ maps to a dense subgroup of the free pro-$p$ group on $k$ generators.

But the free pro-$p$ $k$-generator group surjects a non-solvable linear group for large enough $k$ (say $ker\{ GL_m(\mathbb{Z}_p)\to GL_m(\mathbb{Z}/p)\}$, $m\geq 2$). This group is linear and not virtually solvable, hence has a free subgroup by the Tits alternative. So any inverse limit of the free $k$-generated nilpotent groups will have a free subgroup, hence is not amenable.

Addendum: In fact, we may take $k=2$ above, since any finite-index subgroup of $SL_n(\mathbb{Z})$, $n\geq 3$, contains a finite-index rank 2 subgroup.

Suppose that $\Gamma_n$ is the free nilpotent group on $k$ generators of nilpotency class $n$. Any finite $k$-generated $p$-group ($p$ a prime) is nilpotent, hence the free $k$-generated pro-$p$ group is surjected by the (universal) inverse limit of $\Gamma_n$. Hence any inverse limit of $\Gamma_n$ maps to a dense subgroup of the free pro-$p$ group on $k$ generators.

But the free pro-$p$ $k$-generator group surjects a non-solvable linear group for large enough $k$ (say $ker\{ GL_m(\mathbb{Z}_p)\to GL_m(\mathbb{Z}/p)\}$, $m\geq 2$). This group is linear and not virtually solvable, hence any of its dense subgroups has a free subgroup by the Tits alternative. So any inverse limit of the free $k$-generated nilpotent groups will have a free subgroup, hence is not amenable.

Addendum: In fact, we may take $k=2$ above, since any finite-index subgroup of $SL_n(\mathbb{Z})$, $n\geq 3$, contains a finite-index rank 2 subgroup.

Suppose that $\Gamma_n$ is the free nilpotent group on $k$ generators of nilpotency class $n$. Any finite $k$-generated $p$-group ($p$ a prime) is nilpotent, hence the free pro-$p$ group is surjected by the (universal) inverse limit of $\Gamma_n$. Hence any inverse limit of $\Gamma_n$ maps to a dense subgroup of the free pro-$p$ group on $k$ generators.

But the free pro-$p$ $k$-generator group surjects a non-solvable linear group for large enough $k$ (say $ker\{ GL_m(\mathbb{Z}_p)\to GL_m(\mathbb{Z}/p)\}$). This group is linear and not virtually solvable, hence has a free subgroup by the Tits alternative. So any inverse limit of the free $k$-generated nilpotent groups will have a free subgroup, hence is not amenable.

(this answer seems too simple, so I'm suspecting I'm missing something or made a mistake).

Suppose that $\Gamma_n$ is the free nilpotent group on $k$ generators of nilpotency class $n$. Any finite $k$-generated $p$-group ($p$ a prime) is nilpotent, hence the free $k$-generated pro-$p$ group is surjected by the (universal) inverse limit of $\Gamma_n$. Hence any inverse limit of $\Gamma_n$ maps to a dense subgroup of the free pro-$p$ group on $k$ generators.

But the free pro-$p$ $k$-generator group surjects a non-solvable linear group for large enough $k$ (say $ker\{ GL_m(\mathbb{Z}_p)\to GL_m(\mathbb{Z}/p)\}$, $m\geq 2$). This group is linear and not virtually solvable, hence has a free subgroup by the Tits alternative. So any inverse limit of the free $k$-generated nilpotent groups will have a free subgroup, hence is not amenable.

Addendum: In fact, we may take $k=2$ above, since any finite-index subgroup of $SL_n(\mathbb{Z})$, $n\geq 3$, contains a finite-index rank 2 subgroup.