$\newcommand{\Om}{\Omega}$Let $(Y_n,Z_n):=2^{n/2}(\hat f(y),\hat f(z))$ for distinct $y,z$ in $\Om^n:=\{0,1\}^n$. Then the limit distribution of $(Y_n,Z_n)$ (as $n\to\infty)$ is the standard bivariate normal distribution.
Indeed, for the joint characteristic function $\phi_n$ of $(Y_n,Z_n)$, any any real $s$ and $t$, and $w:=y-z\ne0$ we have \begin{align*} \phi_n(s,t)&=E\exp\big\{i2^{n/2}(s\hat f(y)+t\hat f(z))\big\} \\ &=E\exp\Big\{\frac i{2^{n/2}}\,\sum_{x\in\Om^n}[s(-1)^{x\cdot y}+t(-1)^{x\cdot z}]f(x)\Big\} \\ &=\prod_{x\in\Om^n}E\exp\Big\{\frac i{2^{n/2}}\,[s(-1)^{x\cdot y}+t(-1)^{x\cdot z}]f(x)\Big\} \\ &=\prod_{x\in\Om^n}\cos\Big\{\frac1{2^{n/2}}\,[s(-1)^{x\cdot y}+t(-1)^{x\cdot z}]\Big\} \\ &=\prod_{x\in\Om^n}\cos\Big\{\frac1{2^{n/2}}\,[s+t(-1)^{x\cdot w}]\Big\}. \end{align*} The third equality in the above display holds because the $f(x)$'s are independent, and the fourth equality there holds because $P(f(x)=\pm1)=1/2$ for each $x\in\Om^n$. Since $\ln\cos u=-u^2/2+O(u^4)$ as $u\to0$, we further have \begin{align*} \phi_n(s,t)&=\exp\Big\{-\frac12\frac1{2^n}\,\sum_{x\in\Om^n}\big([s+t(-1)^{x\cdot w}]^2+O(1/2^{2n})\big)\Big\} \\ &=\exp\Big\{-\frac{s^2+t^2}2+O(1/2^n)\Big\}\to e^{-s^2/2}e^{-t^2/2}. \end{align*} Thus, the joint characteristic function $\phi_n$ of $(Y_n,Z_n)$ converges pointwise to the joint characteristic function of the standard bivariate normal distribution -- just as claimed.