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](https://www.researchgate.net/publication/265366840_On_Alspach's_conjecture) 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 decompose $E(K_n)$ 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 two $4$-cycles of $K_n$, so we are done. 

If $n$ is even, we can delete a vertex $v$ and decompose $E(K_{n-1})$ as above.  Then we only need $\frac{n}{2}$ extra $4$-cycles to cover the edges incident to $v$.  This gets us to within $n/ 2$ of the obvious lower bound.