Suppose that $A_i\cap A_j\subseteq A_k$. Then for each $r\in[n]$ one of the following must be true:
- $r\in A_i$ and $r\in A_j$ and $r\in A_k$.
- $r\in A_i$ and $r\notin A_j$, in which case it doesn’t matter whether $r\in A_k$.
- $r\notin A_i$ and $r\in A_j$, in which case it doesn’t matter whether $r\in A_k$.
- $r\notin A_i$ and $\notin A_j$, in which case it doesn’t matter whether $r\in A_k$.
Since each $r\in[m]$ is in any given $A_s$ with probability $\frac12$, the probability of (1) is $\frac12\cdot\frac12\cdot\frac12=\frac18$. The probability of (2) is $\frac12\cdot\frac12\cdot1=\frac14$, and exactly the same calculation gives the probability of (3) and the probability of (4). These four possibilities are mutually exclusive, so the probability that one of them is true is $\frac18+\frac14+\frac14+\frac14=\frac78$.
In other words, for each $r\in[n]$ the probability that $r$ behaves in one of the ways consistent with $A_i\cap A_j\subseteq A_k$ is $\frac78$. This is the case independently for each $r\in[n]$, so the probability that all of them behave properly is $\left(\frac78\right)^n$.