Let $A$ have $n$ elements. The number of sequences of chains
$$S_1 \subset S_2 \subset \dots \subset S_{n-1} \subset A$$
such that $|S_i| = i$ is $n!$: for each permutation $(a_1, a_2, \dots, a_n)$ of elements from $A$ let $S_i = S_{i-1} \cup \{a_i\}$.

Fix a subset $B$ of $A$. If $|B| = k$, then the chains containing $B$
$$S_1 \subset S_2 \subset \dots \subset B \subset \dots S_{n-1} \subset A$$
is $k!(n-k)!$, because there are $k!$ choices for $S_1, \dots, S_{k-1}$ and $(n-k)!$ choices for $S_{k+1}, \dots, S_{n-1}$.
If $B$ and $B'$ are such that $B \not\subset B'$ and $B' \not\subset B$, then every pair of chains, one containing $B$ and other containing $B'$, are disjoint. So if $B_1, B_2, \dots, B_m$ are subsets such that $B_i \not\subset B_j$ and $B_j \not\subset B_i$ then, denoting $|B_i| = m_i$,
$$\sum_{i=1}^{m} m_i!(n - m_i)! \le n! \iff \sum_{i=1}^{n} \frac{1}{\binom{n}{m_i}} \le 1$$
The maximum value of $\binom{n}{m}$ is $\binom{n}{\lfloor \frac{n}{2} \rfloor}$. Hence
$$\sum_{i=1}^{m} \frac{1}{\binom{n}{m_i}} \le 1 \implies \sum_{i=1}^{m} \frac{1}{\binom{n}{\lfloor \frac{n}{2} \rfloor}} \le 1 \iff m \le \binom{n}{\lfloor \frac{n}{2} \rfloor}$$

One example with $m = \binom{n}{\lfloor \frac{n}{2} \rfloor}$ is considering all subsets with $\lfloor \frac{n}{2} \rfloor$ elements.