Strictly sepaking, a composite boson lives in the Fock space of its constituent particles. However, for a sufficiently entangled set of fermions, the corresponding composite particle
\begin{align}
\Lambda^\dagger = \sum_{ij} \lambda_{ij} c^\dagger_i c^\dagger_j, \quad \{c_i, c_j^\dagger\} = \delta_{ij}, \quad \lambda_{ij} = -\lambda_{ji} \in \mathbb{C},
\end{align}
obeys commutation relations that can be approximated as bosonic under certain circumnstances,
\begin{align}
[\Lambda, \Lambda^\dagger] \approx 1.
\end{align}
The effective particle can then be thought as living in some subspace of the fermionic Hilbert space where it can approximately act as a boson. One of these conditions is that the occupation number of the composite boson is small compared with the size of the subspace (i.e. the space where $\lambda_{ij}$ is significant).
One way to see why size of the entanglement space is important is by considering the limiting cases. If we take the fermions to be entantgled over only two modes such that $\lambda_{12} = \lambda/2 = -\lambda_{21}$ and $\lambda_{ij}$ is zero otherwise, the resulting composite particle is $\Lambda^\dagger = \lambda c^\dagger_1 c^\dagger_2$. An important consequence is that this particle obeys an exclusion principle: $(\Lambda^\dagger)^2 = 0$. It therefore cannot possibly be a boson!
On the other hand, if we let $\lambda_{ij} = \lambda/2N = -\lambda_{ji}$ for all $i<j$, with $i,j = 1,...,N$, the same exclusion principle no longer applies
\begin{align}
(\Lambda^\dagger)^2
&= \left(\sum_{ij} \lambda_{ij} c^\dagger_i c^\dagger_j\right) \left(\sum_{kl} \lambda_{kl} c^\dagger_k c^\dagger_l\right) \\
&= \left(\sum_{i<j} \frac{\lambda}{N} c^\dagger_i c^\dagger_j\right) \left(\sum_{k<l} \frac{\lambda}{N} c^\dagger_k c^\dagger_l\right) \\
&= \frac{\lambda^2}{N^2} \sum_{i<j} \sum_{k<l} c^\dagger_i c^\dagger_j c^\dagger_k c^\dagger_l \neq 0.
\end{align}
In fact, we can keep evaluating expressions of the form $(\Lambda^\dagger)^n$, which will lead to terms of order $2n$ in the fermionic operators. We will keep finding "room" for the fermions to occupy in the Hilbert space up until we reach $2n=N$, at which point we have $(\Lambda^\dagger)^N/2 \propto \prod_i^N c_{i}^\dagger$ (assuming an even number of modes). At this point there are no more states that can be occupied, and we reach a new exclusion principle, $(\Lambda^\dagger)^{N/2 + 1} = 0$. This shows that for any finite entangling space defined by the $\lambda_{ij}$, the resulting composite particle will never be, strictly speaking, a boson.
However, as $N \rightarrow \infty$, it becomes increasingly harder to reach this limit. This means that in physical situations the mode correspodning to the composite particle approximately acts as if it can be arbitrarely opccupied, i.e., it's approximately a boson.
