12. For any set $S$, use $|S|$ to denote the number of elements in set $S$, and use $n(S)$ to denote the number of subsets of set $S$. If $A, B, C$ are three finite sets, and satisfy
(1) $|A|=|B|=2016$;
(2) $n(A)+n(B)+n(C)=n(A \cup B \cup C)$.
Then the maximum value of $|A \cap B \cap C|$ is $\qquad$
12. 2015.

From the given conditions, we have
$$
2^{2016}+2^{2016}+2^{|C|}=2^{|A \cup B \cup C|} \text {, }
$$

which simplifies to $1+2^{|C|-2017}=2^{|A \cup B \cup C|-2017}$.
Thus, $|C|=2017,|A \cup B \cup C|=2018$.
Clearly, $|A \cap B \cap C| \leqslant \min \{|A|,|B|\}=2016$.
If $|A \cap B \cap C|=2016$, then $A=B \subseteq C$.
Therefore, $|A \cup B \cup C|=|C|=2017$, which is a contradiction.
Hence, $|A \cap B \cap C| \leqslant 2015$.
Construct: $A=\{1,2, \cdots, 2015,2016\}$,
$$
\begin{array}{l}
B=\{1,2, \cdots, 2015,2017\}, \\
C=\{1,2, \cdots, 2015,2016,2018\}
\end{array}
$$

which satisfies the conditions.
Thus, the maximum value of $|A \cap B \cap C|$ is 2015.