For $k \in \{1, 2, \dots, n\}$, denote by $a_k = \text{card}\{F \in \mathcal{F} \mid k \in F\}$.

Then $a_k \in \{0, 1, \dots, n\}$, and the sum whose maximum we want to estimate is
$$s = \sum_{F \in \mathcal{F}} \text{card}(F) = \sum_{k=1}^{n} a_k.$$
The connection condition implies that if $a_k \neq k$, then $a_{a_k} = k$, for any $k \in \{1, 2, \dots, n\}$. Conversely, any function satisfying this condition corresponds to a family $\mathcal{F}$ verifying the hypothesis.

Therefore, the numbers $a_k$ form pairs of two distinct numbers; otherwise, if $a_i = a_j = a$, then $a$ is connected both to $i$ and to $j$, which would imply $i = j$.
We deduce that
$$s \le n + (n-1) + \dots + 1 = \frac{n(n+1)}{2}.$$
Finally, observe that the maximum $\frac{n(n+1)}{2}$ is attained for any family $\mathcal{F}$ for which $a_{a_k} = k$ for all $k \in \{1, 2, \dots, n\}$; for example, the family $\mathcal{F}_0$ formed by $\{1\}, \{1, 2\}, \dots, \{1, 2, \dots, n\}$.