The Answer is $3n^{2}$ Let $M$ be the maximal value of $f_{AB}(\sigma)+f_{BC}(\sigma)+f_{CD}(\sigma)+f_{DA}(\sigma)$. 1) $M \ge 3n^{2}$ When $\sigma=AA\cdots ABB\cdots BCC \cdots CDD \cdots D$, $f_{AB}(\sigma)+f_{BC}(\sigma)+f_{CD}(\sigma)+f_{DA}(\sigma)=3n^{2}$ obviously. 2) $M \le 3n^{2}$ Let's count the number of pairs $AB,BC,CD,DA$ for each(Not necessarily adjacent), called $"Good Pair"$. For any partial sequence consisting of $4$ words $A,B,C,D$, There are $3$ good pairs at most. Then, $3n^{4}$ good pairs are counted for every partial sequences at most. Moreover, some good pair $AB$ has the number of ways to select $CD$ as $n^{2}$, which it concludes $M \le \frac{3n^{4}}{n^{2}}=3n^{2}$