If $n$ is odd, then the answer is $\lceil \binom{n}{2}/4 \rceil$, which is best possible. This follows from two special cases 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 decompose $E(K_n)$ into $4$-cycles and possibly one extra cycle that is 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, then each vertex has odd degree. Since every $4$-cycle passes through each vertex an even number of times, then for each vertex $v$, there will be at least one edge incident to $v$ that is covered twice. Thus, there will be at least $n/2$ edges that are covered twice. Thus, every covering of $E(K_n)$ by $4$-cycles has size at least $\binom{n}{2}/4+n/8$. We prove that this bound can actually be achieved using the same special cases of Alspach's conjecture.
Namely, for $n$ even, the same two decomposition results hold except with $K_n$ replaced by $K_n$ minus a perfect matching $M$, and $\binom{n}{2}$ replaced with $\frac{n(n-2)}{2}$. Thus, by using the previous argument on $K_n-E(M)$, and then covering pairs of edges in the perfect matching $M$ with $4$-cycles of size $\lceil \binom{n}{4} \rceil+n/8$.