I assume that $c$ is small. Then the minimal answer $(A_n)^{cn}$, where $\sqrt{2}\leqslant \liminf A_n\leqslant e^{1/e}$.
Lower bound. Fix $\varepsilon>0$. We prove that for large $n$ each set $A'$ serves for at most $2^{n-cn/2}$ sets $A$ such that $|A|\geqslant n/2$. Indeed, such $A'$ must have size at least $|A'|\geqslant c|A|\geqslant (c/2)n$, therefore there exist at most $2^{n-cn/2}$ oversets of $A'$. Thus we need at least
$$
2^{n-1}/2^{n-cn/2}=2^{cn/2-1}
$$
different sets $A'$.
Upper bound. For each given positive integer $k=\alpha n$, with $\alpha \leqslant c$, fix some $p\in (0,1)$ and choose each set of size $k$ with probability $p$. What is the probability that some set $A$, $|A|=k/c$ (well, I omit integer parts and so on), does not have any chosen subset of size $k$? It equals
$$
(1-p)^{\binom{k/c}{k}}\leqslant e^{-p\binom{k/c}{k}}\leqslant e^{-p\cdot \exp(n\cdot \alpha H(c)/c)},
$$
where $H(t)=-t\log t-(1-t)\log(1-t)$ is entropy function and I use the well-known bound $\binom{N}{cN}\leqslant e^{H(c)N}$. That is why we naturally take $p=n\exp(-n\cdot \alpha H(c)/c)$. Then the probability that $A$ has no chosen subset is at most $e^{-n}$, and even if we sum up by all possible $A$ we still get at most $(2/e)^n$. How many subsets do we choose? Expectation is $$E:=
p\binom{n}{\alpha n}\leqslant p\cdot e^{n\cdot H(\alpha)}=n\exp\left(n\cdot(H(\alpha)-\alpha H(c)/c)\right).
$$
We have $H(c)=-c\log c+c+O(c^2)$ for small $c$. It follows that
$$
H(\alpha)-\alpha H(c)/c=c\cdot \frac\alpha{c}\cdot \log\frac{c}\alpha+O(c^2)\leqslant c/e+O(c^2),
$$
since $x^{-1}\log x\leqslant 1/e$ for $x=c/\alpha\geqslant 1$. It remains to apply some standard estimate for large deviation from the expectation (Markov inequality is enough) and sum up by all $k$.
This bound $e^{O(cn)}$ is better for small $c$ than $e^{H(c)\cdot n}$ proved by taking all sets of size at most $cn$.
