This is just an answer for those who are curious, and it's the best I can do.
If $A_1 = A_2 = \ldots = A_n$
then
$c_1 = \binom{n}{1} \times |A_1|$
$c_2 = \binom{n}{2} \times |A_1|$
$\vdots$
$c_n = \binom{n}{n} \times |A_1|$
so $c_1, c_2, \ldots, c_n$ is not a decreasing sequence, as the binomial sequence is not a decreasing sequence.
Otherwise if $A_1 \neq A_2 \neq \ldots \neq A_n$
Then, for any element in $A_1 \cap A_2$, there's at least two copies of it in $A_1 + A_2$, so it appears as though $c_1 \geq 2 c_2$ if the elements of $ \{ A_i \cap A_j : 1\leq i<j\leq n \}$ are all disjoint. This argument can be extended to say that $c_1 \geq n c_n$ is guaranteed. And even $c_{n-1} \geq n c_n$ is guaranteed because each element of the $n$-intersection is an element of $n$ amount of $(n-1)$-intersections. But a much weaker statement also follows: that $c_{n-2} \geq \frac{n-1}{n} c_{n-1}$ because despite each element of a $(n-1)$-intersection being elements of $n-1$ amount of $(n-2)$-intersections, there are at least $n$ copies of each element amongst all of the $(n-1)$-intersections, and those elements will not be included $n$ times in each $(n-2)$-intersection.
The takeaway from this line of reasoning I belive, is that $c_{k-2} \geq c_{k-1}$ iff there are at least $\frac{1}{k}$ elements in the $(k-2)$-intersections which are not elements of any $(k-1)$-intersection. Furthermore the sequence $c_1, c_2, \ldots, c_n$ is only decreasing if this is true for all relevant $k$.