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I couldn't resist... A different answerconstruction, which is "irreducible" also for odd $n$ unlike the construction of Ace of Base, is already implied by the very first example the OP gives. (Did you find it by computer and didn't notice?) Look at the differences of the numerators... It can be straight away generalized to $$\boxed{\sin\left(\dfrac{\pi}{2^{n+1}+2} \right)\prod\limits_{k=1}^{n-1}\sin\left(\dfrac{2^n+2^k+1}{2^{n+1}+2}\pi \right)=\dfrac1{2^n}}.$$ Sure enough, denoting $\sin\left(\dfrac{k\pi}{2^{n+1}+2}\right)$ and $\cos\left(\dfrac{k\pi}{2^{n+1}+2}\right)$ by $s_k$ and $c_k$ respectively, we have $$LHS=s_1\prod\limits_{k=1}^{n-1}c_{2^k} =\frac1{2c_1}s_2\prod\limits_{k=1}^{n-1}c_{2^k} =\frac1{4c_1}s_4\prod\limits_{k=2}^{n-1}c_{2^k} =\cdots=\frac{s_{2^{n}}}{2^nc_1}=\frac1{2^n}.$$
I couldn't resist... A different answer, which is "irreducible" also for odd $n$ unlike the construction of Ace of Base, is already implied by the very first example the OP gives. (Did you find it by computer and didn't notice?) Look at the differences of the numerators... It can be straight away generalized to $$\boxed{\sin\left(\dfrac{\pi}{2^{n+1}+2} \right)\prod\limits_{k=1}^{n-1}\sin\left(\dfrac{2^n+2^k+1}{2^{n+1}+2}\pi \right)=\dfrac1{2^n}}.$$ Sure enough, denoting $\sin\left(\dfrac{k\pi}{2^{n+1}+2}\right)$ and $\cos\left(\dfrac{k\pi}{2^{n+1}+2}\right)$ by $s_k$ and $c_k$ respectively, we have $$LHS=s_1\prod\limits_{k=1}^{n-1}c_{2^k} =\frac1{2c_1}s_2\prod\limits_{k=1}^{n-1}c_{2^k} =\frac1{4c_1}s_4\prod\limits_{k=2}^{n-1}c_{2^k} =\cdots=\frac{s_{2^{n}}}{2^nc_1}=\frac1{2^n}.$$
A different answer, which is "irreducible" also for odd $n$ unlike the construction of Ace of Base, is already implied by the very first example the OP gives. (Did you find it by computer and didn't notice?) Look at the differences of the numerators... It can be straight away generalized to $$\boxed{\sin\left(\dfrac{\pi}{2^{n+1}+2} \right)\prod\limits_{k=1}^{n-1}\sin\left(\dfrac{2^n+2^k+1}{2^{n+1}+2}\pi \right)=\dfrac1{2^n}}.$$ Sure enough, denoting $\sin\left(\dfrac{k\pi}{2^{n+1}+2}\right)$ and $\cos\left(\dfrac{k\pi}{2^{n+1}+2}\right)$ by $s_k$ and $c_k$ respectively, we have $$LHS=s_1\prod\limits_{k=1}^{n-1}c_{2^k} =\frac1{2c_1}s_2\prod\limits_{k=1}^{n-1}c_{2^k} =\frac1{4c_1}s_4\prod\limits_{k=2}^{n-1}c_{2^k} =\cdots=\frac{s_{2^{n}}}{2^nc_1}=\frac1{2^n}.$$