The answer is $\textstyle\binom{n}{\lfloor n/2\rfloor}$, and attained by taking all such subsets. It suffices to show $|A|\le \textstyle\binom{n}{\lfloor n/2\rfloor}$.This is the classical Lubell-Yamamoto-Meshalkin inequality. Here's a proof based on the probabilistic method. First, note that we are done if $\varnothing \in A$, as $|A|=1$ in this case. For any $S\in A$, let $E_S$ be the event that a randomly chosen permutation starts with the elements of $S$. That is, \[ E_S = \{\{\sigma(1),\dots,\sigma(S)\}=S\}, \]where $\sigma$ is taken uniformly at random among all $n!$ permutations. Note that since none of the elements of $A$ contain another subset from $A$, we have that the events $(E_S:S\in A)$ are disjoint. So, \[ 1\ge \mathbb{P}\bigl[\bigcup_{S\in A}E_S\bigr] = \sum_{S\in A}\mathbb{P}[E_S] = \sum_{S\in A}\frac{|S|! (n-|S|)!}{n!} =\sum_{S\in A} \frac{1}{\binom{n}{|S|}}. \]Using the fact $\textstyle\binom{n}{k}\le \binom{n}{\lfloor n/2\rfloor}$ valid for all $0\le k\le n$, we conclude $|A|\le \textstyle \binom{n}{\lfloor n/2\rfloor}$.