The idea is that in every circle configuration we can divide the circles into $\bar{n}$ of those that generate only double intersection points(call them "black") and $\tilde{n}$ of those that we thread through the double intersection points to generate the triple intersection points (call them colored) , $\bar{n} + \tilde{n} = n$. Any $\bar{n}$ black circles can generate at most $\bar{n} (\bar{n}-1)$ double intersection out of which at most $\bar{n}$ intersection points can be placed on a single colored circle different from any black circles because if you choose any $\bar{n}+1$ double intersection points there will always be 3 points from the same circle. Therefore, any colored circle can generate at most $\bar{n}$ intersection points and the total number of intersection points cannot be larger than $\bar{n}\tilde{n}$.

So far in our circle configuration we want to choose $\bar{n}$ and $\tilde{n}$ so that their product was maximal and at the same time keeping $\tilde{n} \leq (\bar{n}-1)$ since we cannot have more than $\bar{n} (\bar{n}-1)$ triple intersection points points.

Considering that $\tilde{n} = n - \bar{n}$ we have an optimization problem to find the right number of black and colored circles: $$ maximize\quad \bar{n}(n-\bar{n}) $$ $$ s.t\quad n-\bar{n} \leq (\bar{n}-1) $$ We can construct the configuration having exactly $\bar{n}(n-\bar{n})$ triple intersections in the following way: Place the black circles symmetrically on the corners of a regular $\bar{n}$-gon such that the diameter of the circles is the size of an edge of the $\bar{n}$-gon, this will generate $\bar{n}-1$ layers of $\bar{n}$ double intersection points through which you can thread the remaining $\tilde{n}$ colored circles obtaining exactly the largest possible number of triple intersection points $\bar{n}(n-\bar{n})$.

The following construction shows that $c_3(n)\ge n^2/4-1$. Put $k=\lceil \frac{n+1}2\rceil$ and draw a regular $k$-gon $K$ with side $1$. Initially, we place $k$ black unit circles centered in the corners of $K$. This generates $k - 1$ circular layers of $k$ double intersection points (see, for instance, the last picture at the question). Since $k-1\ge n-k$, we can cover $n-k$ layers by red circles, achieving at least $k$ triple intersection points on each layer and $k(n-k)\ge n^2/4-1$ of them in total.