TL;DR
Quantum capacity of $\mathcal{N}_2\circ\mathcal{N}_1$ can be anywhere between zero and the minimum of the quantum capacities of $\mathcal{N}_1$ and $\mathcal{N}_2$.
Background
Quantum capacity of a quantum channel $\mathcal{N}$ is defined as the greatest real number $Q(\mathcal{N})$ such that for any $R < Q(\mathcal{N})$ (representing a transmission rate) and any $\delta > 0$ (representing an acceptable error rate), there exists a quantum error correcting code with encoding operation $\mathcal{E}$ that maps $m$ qubits into $n$ inputs to the channel $\mathcal{N}$ and decoding operation $\mathcal{D}$ that maps $n$ outputs from the channel $\mathcal{N}$ to $m$ qubits which enables recovery of any $m$-qubit input state $|\psi\rangle$ with fidelity $1-\delta$ and such that $\frac{m}{n} > R$. See also discussion on page $1$ and figure $1$ on page $4$ in this paper.
Let $\mathcal{N}_2\circ\mathcal{N}_1$ denote the channel representing the application of $\mathcal{N}_1$ followed by the application of $\mathcal{N}_2$ and note that
$$
0 \le Q(\mathcal{N}_2\circ\mathcal{N}_1) \le \min(Q(\mathcal{N}_1),Q(\mathcal{N}_2)).\tag1
$$
The first inequality follows from the fact $m$ and $n$ in the definition above are non-negative integers and therefore $Q(\mathcal{N})$ is a non-negative real number for any channel $\mathcal{N}$. The second inequality follows from the fact that we can absorb $\mathcal{N}_1$ (respectively, $\mathcal{N}_2$) into the encoding operation $\mathcal{E}$ (respectively, the decoding operation $\mathcal{D}$) to improve $Q(\mathcal{N}_2)$ (respectively, $Q(\mathcal{N}_1)$) to at least $Q(\mathcal{N}_2\circ\mathcal{N}_1)$.
Two extreme cases
Can we tighten the bounds in $(1)$? It turns out that without additional assumptions on the channels, we can't. To show this, we will use the quantum erasure channel
$$
\mathcal{R}_\epsilon(\rho) = (1 - \epsilon)\rho + \epsilon|2\rangle\langle 2|
$$
to construct two examples, one saturating the left inequality in $(1)$ and one saturating the right inequality. We choose the erasure channel because its quantum capacity is known and is given by the formula
$$
Q(\mathcal{R}_\epsilon) = \max(0, 1 - 2\epsilon),\tag2
$$
see equation $(2)$ on page $2$ (and figure $2(a)$ on page $4$) in this paper. Now, the effect of composing two instances of the erasure channel is
$$
\begin{align}
\mathcal{R}_\epsilon(\mathcal{R}_\epsilon(\rho)) &= (1 - \epsilon)\mathcal{R}_\epsilon(\rho) + \epsilon\mathcal{R}_\epsilon(|2\rangle\langle 2|) \\
&= (1 - \epsilon)^2\rho + (2\epsilon - \epsilon^2)|2\rangle\langle 2| \\
&= \mathcal{R}_{2\epsilon-\epsilon^2}(\rho).
\end{align}
$$
Thus,
$$
\begin{align}
0 &\le Q(\mathcal{R}_{1/3} \circ \mathcal{R}_{1/3}) \le \min(Q(\mathcal{R}_{1/3}), Q(\mathcal{R}_{1/3})) \\
0 &\le Q(\mathcal{R}_{5/9}) \le \min\left(\frac13, \frac13\right) \\
0 &\le 0 \le \frac13 \\
\end{align}
$$
saturating the left inequality in $(1)$ and
$$
\begin{align}
0 &\le Q(\mathcal{R}_0 \circ \mathcal{R}_0) \le \min(Q(\mathcal{R}_0), Q(\mathcal{R}_0)) \\
0 &\le Q(\mathcal{R}_0) \le \min\left(1, 1\right) \\
0 &\le 1 \le 1 \\
\end{align}
$$
saturating the right inequality in $(1)$.
