Truth set of a Universal Quantifier and Family of sets 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 $x$. Assume $x\in A$ and $A\notin F$, doesn't that mean that $x$ is in $\cap F$(including any element $x\in A$ for any $A\notin F$)   ? So any $x$ can be in $\cap F$ if $A\notin F$ since $\forall A(A\in F \implies x\in A)$ will be true if $A\notin F$. 
This case is similar to $A=\{z|\forall n (n\in N\implies z=n)\}$($N$ is the set of natural numbers), if $n\notin N $  then the set $A$ can contain any value in the real system since $\forall n (n\in N\implies z=n)$ will be true if $n\notin N$.
 A: If there is some $A\notin F$, then we know nothing about whether $x\in A$ or not. 
When the antecedent of an implication is false, we know the entire implication, as a whole, is true; but we no nothing about the truth value of its conclusion $(\text{e.g., whether }x \in A)$.
A: Logic notation offers brevity but can cause confusion (as in the intuition of your question).
For my response, I will avoid logic notation (but keep set notation), and hopefully you will understand what the condition $\forall A[A\in F\Rightarrow x\in A]$ which you call "elementhood test" truly states.
Elementhood Test
$x\in \bigcap F$
if and only if
for each set $A$, if $A\in F$, then $x\in A$.
Your Question
Let $\mathcal{A}$ be a set such that $x\in \mathcal{A}$ and $\mathcal{A}\notin F$. "Doesn't that mean $x\in\bigcap F$?"
Answer
No.
Explanation
In order for $x$ to be an element of $\bigcap F$, $x$ must pass the elementhood test. We know there exists a set $\mathcal{A}$ such that $x\in\mathcal{A}$ and $\mathcal{A}\notin F$. We do not know that for each set $\mathsf{A}$, if $\mathsf{A}\in F$, then $x\in\mathsf{A}$ (elementhood test).
Alternative Explanation (using logic notation)
You are confusing $$\exists A[A\in F\Rightarrow x\in A]$$ with $$\forall A[A\in F\Rightarrow x\in A].$$That is, the two statements are not the same. Your question can be rewritten as
$$(\exists A[A\in F\Rightarrow x\in A]\implies x\in A)?$$which is why the answer is "No."
A: The logical notation is clear.
The definition of $\cap \mathcal F$ is equivalent to :

$∀a(a \in \mathcal F \rightarrow x \in a) \leftrightarrow x \in \cap \mathcal F$.

The formula must be read as universally quantified, i.e. as :

for all $x$, if the "defining condition" : $∀a(a \in \mathcal F \rightarrow x \in a)$ applies, then we can conclude that $x \in \cap \mathcal F$, and vice versa.

But also the defining condition is universally quantified, and we must remember that in "standard" set theory, like $\mathsf {ZF}$, all the objects in the domain are sets.
Thus, consider a simple example with $\mathcal F = \{ \{ 1, 2 \}, \{ 2, 3 \} \}$.
If we "calculate" it, we have that with $x := 1$ and $a := \{ 1, 2 \}$ we have that $a \in \mathcal F \rightarrow x \in a$ is true, because both the antecedent and the consequent of the conditional are true.
But with $x := 1$ and $a := \{ 2, 3 \}$ we have that now $a \in \mathcal F \rightarrow x \in a$ is false, because the antecedent is true but the consequent is false.
Thus, having found for $x := 1$ an $a$ such that $a \in \mathcal F \rightarrow x \in a$ is false, we conclude that $\forall a(a \in \mathcal F \rightarrow x \in a)$ is false and thus $1 \notin \cap \mathcal F$. 
The same applies with $x := 3$.
Now consider $x := 2$; in both cases $a := \{ 1, 2 \}$ and $a := \{ 2, 3 \}$, we have that $a \in \mathcal F \rightarrow x \in a$ is true.
But what happens with all other $a$ of the "universe" ? We have that for all $a$ except the two above, $a \notin \mathcal F$; thus, $a \in \mathcal F \rightarrow x \in a$ is vacuously true for them.
Thus, in all cases : $\forall a(a \in \mathcal F \rightarrow x \in a)$ is true, and we conclude with :

for $x := 2$, we have that $x \in \cap \mathcal F$.

The "tricky" case is when $\mathcal F$ is empty; see Herbert Enderton, Elements of Set Theory (1977), page 25 :

There is one troublesome extreme case. What happens if $\mathcal F$ is empty ? For any $x$ at all, it is vacuously true that $x$ belongs to every member of $\emptyset$. (There can be no member of $\emptyset$ to which $x$ fails to belong.) Thus it looks as if $\cap \emptyset$ should be the class $V$ of all sets. 

But we have that in a theory like $\mathsf {ZF}$ the class $V$ is not a set.

The situation is analogous to division by zero in arithmetic. How does one define $a/0$ ? One option is to leave $\cap \emptyset$ undefined, since there is no very satisfactory way of defining it. [...] The other option is to select some arbitrary scapegoat (the set $\emptyset$ is always used for this) and define $\cap \emptyset$ to equal that object.

Note. See also Patrick Suppes, Axiomatic set theory (1960) , page 40 :

Theorem 67. $x \in \cap A \leftrightarrow (\forall B)(B \in A \rightarrow x \in B) \land (\exists B)(B \in A)$. 

