Combinatorial discrepancy of the system of all cuts
I doubt you can do better than $O(n^{3/2})$. Here is a lower bound, using old arguments by Spencer and Erdos, that shows that the discrepancy is $\Omega(n^{3/2})$ if you expand the set system to cuts of induced subgraphs. See the edit for a lower bound for cuts as well, suggested by Domotor.

Let $V$ be the vertex set of the complete graph, and fix a function $f:{V \choose 2} \to \{-1, 1\}$. Split $V$ into $L$ and $R$, each of equal size. Let $S$ be a random subset of $L$. Then, for any $v \in R$, 
$$
\mathbb{P}\left(\Bigl|\sum_{u \in S}{f(u, v)}\Bigr| \geq c\sqrt{n}\right) \ge 1/4,
$$
for a suitable constant $c$. (This can be proved using the Central Limit Theorem or a 4-th moment argument.) Letting $T_0$ be the subset of all $v \in R$ satisfying $\Bigl|\sum_{u \in S}{f(u, v)}\Bigr| \geq c\sqrt{n}$, we have that $\mathbb{E}|T_0| \geq n/8$. So there exists some choice of $S$ such that $T_0$ has size at least $n/8$: fix such an $S$ and pick $T = \{v \in R: \sum_{u \in S}{f(u, v)} \geq c\sqrt{n}\}$ or $T = \{v \in R: \sum_{u \in S}{f(u, v)}\leq -c\sqrt{n}\}$, whichever is larger. Clearly 
$$
\Bigl|\sum_{u \in S, v \in T}{f(u,v)}\Bigr| \ge \frac{cn^{3/2}}{16}.
$$

EDIT: As Domotor pointed out in a comment the above suffices for a $\Omega(n^{3/2})$ lower bound on the discrepancy of cuts as well. Let $S$ and $T$ be the sets above, and $U = V \setminus (S \cup T)$. Define $f(X,Y) = \sum_{x \in X, y \in Y}{f(x,y)}$ for any two subsets $X,Y\subseteq V$. Assume without loss of generality that $f(S,T) \ge c'n^{3/2}$ for $c' = c/16$ (the case $f(S,T) \le -c'n^{3/2}$ is analogous). Then at least one of the cases below holds:


$f(S,U) \ge -c'n^{3/2}/2$ and $f(S, T \cup U) \ge c'n^{3/2}/2$; 
$f(T,U) \ge -c'n^{3/2}/2$ and $f(S\cup U, T) \ge c'n^{3/2}/2$;
$f(S \cup T, U) = f(S, U) + f(T,U) &lt; -c'n^{3/2}$.


In each case the discrepancy of some cut is $\Omega(n^{3/2})$.