Let $S$ be the complement of $A_1 \cup A_2 \cup \dots \cup A_{11}$ in $A$; we wish to prove that $|S| \le 60$. For $\ell \ge 0$, define
$$\theta(\ell) = \left(1 - \frac{\ell}{2}\right) \left(1 - \frac{\ell}{3}\right) = 1 - \frac{2}{3}\ell + \frac{1}{3}\binom{\ell}{2}.$$
Note that $\theta(0) = 1$ and $\theta(\ell) \ge 0$ for any integer $\ell > 0$. For $n \in A$, let $\ell(n)$ be the number of sets among $A_1, \dots, A_{11}$ containing $n$. Since $S$ is the intersection of the complements of the $A_i$, we see that
$$|S| \le \sum_{n \in A} \theta(\ell(n)).$$
On the other hand, we have
$$\sum_{n \in A} \theta(\ell(n)) = \sum_{n \in A} \left(1 - \frac{2}{3}\ell(n) + \frac{1}{3}\binom{\ell(n)}{2}\right) = |A| - \frac{2}{3}\sum_i |A_i| + \frac{1}{3}\sum_{i<j} |A_i \cap A_j|.$$
Putting these two equations together, we obtain
$$|S| \le 225 - \frac{2}{3} \cdot 11 \cdot 45 + \frac{1}{3} \cdot \binom{11}{2} \cdot 9 = 60,$$
and therefore $|A_1 \cup A_2 \cup \dots \cup A_{11}| \ge 165$.

It remains to give an example showing that this lower bound is best possible. Let $p_1, p_2, \dots, p_{11}$ be a set of $11$ distinct primes, and let $A'$ denote the set of all products of three of these primes. Let $A'' = \{q_1, q_2, q_3, \dots, q_{60}\}$ be a set of $60$ distinct positive integers that are all coprime to $p_1, \dots, p_{11}$. Set $A = A' \cup A''$, and define
$$A_i = \{n \in A' : p_i \mid n\}.$$
Then $|A_i| = \binom{10}{2} = 45$, $|A_i \cap A_j| = \binom{9}{1} = 9$, and
$$|A_1 \cup A_2 \cup \dots \cup A_{11}| = |A'| = \binom{11}{3} = 165.$$
Finally, $|A| = |A'| + |A''| = 165 + 60 = 225$, so this $A$ and $A_1, \dots, A_{11}$ give a valid example.