I know there are answers to this specific question, but I'm trying to solve it through a specific line of thought which was used in a textbook on modern algebra for the case with 2 sets. It goes like this:
For two finite sets, let $A'$ and $B'$ be the parts of $A$ and $B$ which have elements that are NOT in $A \cap B$. The number of elements of the union of the two sets, denoted by $n ( A \cup B )$, can be found by:
$ n(A \cup B) = n(A') + n(B') + n(A \cap B)$
But we have: $n(A') = n(A) - n(A \cap B)$ and $n(B') = n(B) - n(A \cap B)$. Those two relations plugged back in the first expression give us the correct answer:
$n(A \cup B) = n(A) + n(B) - n(A \cap B)$
Now to the case with 3 sets $A$, $B$ and $C$. I followed the same reasoning, taking the parts $A'$, $B'$ and $C'$ that are not in any of the intersections and added them up:
$ n(A \cup B \cup C) = n(A') + n(B') + n(C') + n(A \cap B) + n(A \cap C) + n(B \cap C) + n(A \cap B \cap C)$
Where we have:
$n(A') = n(A) - n(A \cap B) - n(A \cap C) - n(A \cap B \cap C)$
$n(B') = n(B) - n(A \cap B) - n(B \cap C) - n(A \cap B \cap C)$
$n(C') = n(C) - n(A \cap C) - n(B \cap C) - n(A \cap B \cap C)$
Plugging it back those relations I find an incorrect answer with a factor of $-2$ multiplying $n (A \cap B \cap C)$, that is:
$n(A \cup B \cup C ) = n(A) + n(B) + n(C) - n(A \cap B) - n(A \cap C) - n(B \cap C) - 2 n (A \cap B \cap C)$
Can someone point out the mistake in this reasoning? I'd appreciate very much.