Chernoff Bounds for settings with limited dependence
If $X_1, \cdots, X_n$ are $k$-wise independent random variables on $[0,1]$, then $$\text{Pr} \left[ \left| \sum_i X_i - \text{Ex}\left[ \sum_i X_i \right] \right| \geq n \cdot \varepsilon \right] \leq \left( \frac{k^2}{4 n \varepsilon^2} \right)^{\lfloor k/2 \rfloor}.$$
Note that this bound does not necessarily improve with greater $k$. So you may want to choose a smaller $k$ to get the best results.

Proof Sketch. W.L.O.G. $k$ is even. Consider the $k^\text{th}$ moment: $$\text{Ex} \left[ \left( \sum_i X_i - \text{Ex}\left[ \sum_i X_i \right] \right)^k \right] = \sum_{S \in [n]^k} \text{Ex} \left[ \prod_{i \in S} X_i - \text{Ex}[X_i] \right].$$
Each summand on the RHS is nonzero only if every $i$ appears at least twice in $S$. (Otherwise, since $\text{Ex}[X_i-\text{Ex}[X_i]]=0$ and the other terms are independent, the product is zero.) So each nonzero summand has at most $k/2$ different values of $i$. So there are at most ${n \choose k/2} (k/2)^k$ nonzero summands. Each summand is bounded by $1$. Thus $$\text{Ex} \left[ \left( \sum_i X_i - \text{Ex}\left[ \sum_i X_i \right] \right)^k \right] \leq {n \choose k/2} (k/2)^k \leq (n \cdot k^2 /4)^{k/2}.$$ Now apply Markov's inequality to obtain the result. Q.E.D. (Sorry the proof sketch is so brief. I can elaborate if you want.)

This appears as Prob. 3.8 p. 52 in Pseudorandomness by S. Vadhan.