I don't know your purpose, but here's some not-probably-sharp estimate that works for any $k$. Put $\rho=\frac{1}{d}\|A|_{({\mathbb C}1)^\perp}\|$ and $\gamma$ to be the positive root of $t^2-\rho t -\rho=0$. One has $\gamma<\sqrt{2\rho}<1$ when $\rho<\frac{1}{2}$. Then for any $f$ with $\sum f(v)=0$ and $\|f\|_\infty\le1$, one has $$\frac{1}{|V|\cdot d^{k-1}}\left|\sum_{v_1,v_2,\ldots,v_k : \{v_i,v_j\}\in E} f(v_1)\cdots f(v_k)\right| \le \gamma^k.$$$$\frac{1}{|V|\cdot d^{k-1}}\left|\sum_{v_1,v_2,\ldots,v_k : \{v_i,v_{i+1}\}\in E} f(v_1)\cdots f(v_k)\right| \le \gamma^k.$$
Proof. For $D:=\mathrm{diag}\,f \in B(\ell_2V)$ and $B:=\frac{1}{d}AD$, the LHS is $\frac{1}{|V|}|\langle B^{k-1}1_V,f\rangle|$. With respect to the orthogonal decomposition $\ell_2V={\mathbb C}1_V\oplus ({\mathbb C}1_V)^\perp$, one writes $B$ as an operator matrix $B=\left[\begin{smallmatrix} 0 & b \\ c & d \end{smallmatrix}\right]$, where $\| b\|\le 1$, $\|c\|\le\rho$, and $\|d\|\le\rho$. Hence for $C:=\left[\begin{smallmatrix} 0 & 1 \\ \rho & \rho \end{smallmatrix}\right] \in M_2({\mathbb R})$ with the eigenvalue $\gamma>0$ and the eigenvector $\left[\begin{smallmatrix} 1 \\ \gamma \end{smallmatrix}\right]$, one gets $$\frac{1}{|V|}|\langle B^{k-1}1_V,f\rangle| \le \left[\begin{smallmatrix} 0 & 1 \end{smallmatrix}\right] C^{k-1} \left[\begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\right] \le \left[\begin{smallmatrix} 0 & 1 \end{smallmatrix}\right] C^{k-1} \left[\begin{smallmatrix} 1 \\ \gamma \end{smallmatrix}\right]=\gamma^k.$$
It's probably worth noting that the same proof shows $$\frac{1}{|V|}\sum_{v_1\in V}\left|\frac{1}{d^{k-1}}\sum_{v_2,\ldots,v_k : \{v_i,v_{i+1}\}\in E} f_1(v_1)\cdots f_k(v_k)\right|^2 \le 2\gamma^{2(k-1)}.$$