We will count each element in $\frac{7}{8}$ of the sets so total amount will be $7n*8^{n-1}$.

My solution: Let L be a set of $l$ elements $(a_1,a_2,\ldots,a_l)$, So we make that $L=\mid A \cap B \cap C\mid$ We need to choose any of $n-l$ elements that any of them are not in $L$, so they have $2^3-1$ posibilities because any of them can't be an element of $A, B$, and $C$ in the same time, so we have $7^{n-l}$posibilities for $A,B,C$ when L has $l $ elements, we have $\binom{n}{l}$ we will iterate for $l = 0 $ to $l=n$, which one aports \[ 7^{n-l}l\binom{n}{l}\]for all subset that has l elements. Finally We have that \[\sum_{\text{for all}}\mid A \cap B \cap C\mid = \sum_{l=0}^n7^{n-l}l\binom{n}{l}\] Also is well-know $l \binom{n}{l}\equiv 0 \mod{n}$ so \[\sum_{l=0}^n7^{n-l}l\binom{n}{l}\equiv 0\mod{n}\]And we're done!!!!!!!!!!!!!!!!!!!