Expanding on my earlier comment, the concentration has to be really quite good: at least exponential in $N$, the order of the additive group. (Edit: and you can't get better concentration because the state space is only exponentially large.)

Recall the following variant of the Hoeffding-Azuma martingale concentration inequality. Let $\Omega = \Omega_1 \times \cdots \times \Omega_n$ be a product space, and let $X$ be a random variable on $\Omega$ such that $X(\omega)$ changes by at most $k$ if we change a single coordinate of $\omega$. Then $$\mathbb{P}(X \geq (1+t)\mu) \leq \exp (-2t^2\mu^2/nk^2),$$ where $\mu=\mathbb{E}(X)$.

Apply this with $n=2N$, $\Omega$ the space of coin flips determining membership of each element of the group in $A$ and $B$, and $X=E(A,B)$. We have $\mu = N^3/16$ and $k = O(N^2)$ (as each element of the group is in $O(N^2)$ additive quadruples), which gives $$\mathbb{P}(X \geq (1+t)\mu) \leq \exp(-ct^2N),$$ for some absolute constant $c$.