If $n$ is odd, then the answer is $\lceil \binom{n}{2}/4 \rceil$, which is best possible. This follows from a special case of a more general conjecture by Alspach.
For our purposes, we can use a theorem of Heinrich, Horák, and Rosa which says that if $n$ is odd and $a,b,c$ are such that $3a+4b+6c=\binom{n}{2}$, then $E(K_n)$ can be partitioned into $a$ $3$-cycles, $b$ $4$-cycles, and $c$ $6$-cycles. They also proved the same theorem with $3,4,6$ replaced by $3,4,5$.
Thus, if $n$ is odd, it is always possible to decomposed $E(K_n)$ into $4$-cycles and one extra cycle that is possibly a $3$-cycle, a $5$-cycle or a $6$-cycle. The edge set of the extra cycle can obviously be covered with $2$ $4$-cycles of $K_n$, so we are done.