The definition of an intersection of family($F$) is:
$$\cap F=\bigl\{x\mid\forall A: (A\in F \implies x\in A)\bigr\} $$
If my understanding serves me correctly this notation means that all the $x$ that pass the elementhood test are in the set. So $\forall A: (A\in F \implies x\in A)$ must be either true or false for each different $x$. But in this case isn't $F$ also a free variable, so we can't assign a truth value to $\forall A:(A\in F \implies x\in A)$ as it contains a free variable $F$ ($x$ and $A$ are bounded). Is there some implicit assumption about $F$ ? Also in case of an indexed family $F=\{A_i\mid i\in I\}= \bigl\{x\mid\exists i\in I: x=A_i\bigr\}$, is $A_i$ (which is a function of $i$) bound or free (if free isn't it the same problem as the first question) ?
$F$ is not a free variable. It denotes a particular family of sets which you're taking the intersection of. For instance, if $F = \{\{1, 2\}, \{1, 3\}\}$ then your definition says that $\cap F$ consists of those $x$ which are in every set which is in $F$. In my example, the only such $x$ is $1$. Thus the statement $\forall A(A \in F \implies x \in A)$ is true for $1$, but false for $2$ and $3$.
The indexed family just gives a way of listing an infinite set of sets. Namely, it says that for each element $i$ of the set $I$, we have a set $A_i$ which is in $F$ (as an element, not a subset). If one assumes that $A_i \neq A_j$ whenever $i \neq j$, then this implies that the number of sets in $F$ has the same cardinality as $I$.