In your example, algebraic manipulation gives
$$
2^{-M} \sum_{n=0}^{2^M-1} X_n^{(M)} = \prod_{j=0}^{M-1} \cos t h_j.
$$ 
If the $h_j$'s are linearly independent over $\Bbb Q$, then, as you point out, the random variables $\cos t h_j$ approach independence as $t$ is chosen over larger and larger intervals.  Therefore, in the limit, 
$$
Y \sim Z_1 ... Z_M, \qquad Z_i = \cos W_i, \qquad W_1,...,W_M {\rm\ \ i.i.d.\ uniform\ on\ } [0,2\pi).
$$
By symmetry, ${\bf E}[Y]=0$, and since ${\bf E}[Z_i^2]=1/2$ for each $i$, we have ${\bf Var}[Y]=2^{-M}$.  This may explain the result of your numerical experiments.  However, $Y$ does not approach normal after rescaling: if $Y$ is rescaled to unit variance by setting $Y'_M:=2^{M/2} Y$, then $\log |Y'_M|=\log(\sqrt{2} |Z_1|)+...+\log(\sqrt{2} |Z_M|)$, so, since ${\bf E}[\log(\sqrt{2} |Z_i|)]<0$ and ${\bf Var}[\log(\sqrt{2} |Z_i|)]<\infty$,  we can apply the CLT to $\log |Y'_M|$ to show that, as $M\to\infty$, $\log |Y'_M|$ will converge weakly to normal after rescaling, but $Y'_M$ converges weakly to 0.