$\def\ZZ{\mathbb{Z}}\def\RR{\mathbb{R}}$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 $(\RR/\ZZ)^{3}$. 

<b>Proof of pairwise equidsitribution:</b> 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][1], since it is impossible that $a+b=2b=0$ for $(a,b) \neq (0,0)$.
$\square$

<b>Proof of triple non-equidsitribution:</b> We have $(\theta (n+2)^2) - 2 \theta (n+1)^2 + \theta n^2 = 2 \theta$. So 
$$r_{n+2} - 2 r_{n+1} + r_n = 2\theta \bmod \ZZ.$$
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.



  [1]: https://math.stackexchange.com/questions/128/how-do-you-prove-that-pn-xi-for-xi-irrational-and-p-a-polynomial-is-un