$\def\ZZ{\mathbb{Z}}\def\RR{\mathbb{R}}$Here is a suggestion, with some details missing. Let $\theta$ be an irrational number and set $r_n = \{ \theta n^2 \}$, where $\{ \alpha \}$ is the fractional part of $\alpha$. Then I claim that the pairs $(r_n, r_{n+j})$ are equidistributed in $(\RR/\ZZ)^2$, but $(r_n, r_{n+1}, r_{n+2})$ is not equidistributed in $(\RR/\ZZ)^{3}$.
Proof of pairwise equidsitribution: To see that $(r_n, r_{n+j})$ is $0$, we just need to check that the expected value of $\exp((2 \pi i) (a r_n + b r_{n+j})$ is nonzero for any $(a,b) \in \ZZ^2 \setminus \{ (0,0) \}$. In other words, we need the expected values of $\exp((2 \pi i) \theta ((a+b) n^2 + 2bj n + b j^2))$ to be $0$. This follows from Weyl's equidistribution theorem, since it is impossible that $a+b=2b=0$ for $(a,b) \neq (0,0)$. $\square$
Proof of triple non-equidsitribution: We have $(\theta (n+2)^2) - 2 \theta (n+1)^2 + \theta n^2 = 2 \theta$. So $$r_{n+2} - 2 r_{n+1} + r_n = \theta \bmod \ZZ.$$ We also know that $$-2 \leq r_{n+2} - 2 r_{n+1} + r_n \leq 2.$$ So $r_{n+2} - 2 r_{n+1} + r_n$ lies on $4$ hyperplanes through the cube $[0,1)^3$. $\square$
Now, choose any function $f : \RR/\ZZ \to \{ \pm 1 \}$ with average value $0$. Then $f(r_n)$ will have all pointwise correlations $0$, but there is no reason that the triple correlation should vanish.
It probably wouldn't be hard to explicitly show that the triple correlation doesn't vanish, for some well chosen $f$, but I'll stop here.