The product does not tend to the limit zero. For any irrational number $\alpha$ one can show that $$ \limsup_{N\to \infty} \prod_{n=1}^{N} |1- e(n\alpha)| = \infty. \tag{1} $$ (Here I use the usual notation $e(\alpha) =e^{2\pi i\alpha}$, and the product in the question is stated for $\alpha/2$ rather than $\alpha$).

I'll prove a slightly weaker result; namely I'll show that (1) holds for any irrational $\alpha$ for which there are infinitely many rational approximations $a/q$ with $(a,q)=1$ and $$ \alpha=\frac{a}{q} +\beta, \qquad \text{with}\qquad |\beta| \le \frac{1}{100q^2}. \tag{2} $$ The $1/100$ is just chosen for ease of exposition, and a more careful argument can make do with $1/\sqrt{5}$ (any constant $<1/2$ is enough), and then it is well known that every irrational number admits infinitely many approximations with $|\alpha -a/q| \le 1/(\sqrt{5}q^2)$. Condition 2 holds for almost all irrational numbers -- it fails only if the continued fraction expansion only uses numbers below $100$.

Suppose then that $q$ is such that (2) holds, and consider the product in question at $N=q-1$. By the triangle inequality, for $1\le n\le N$, $$ |1-e(n\alpha)| = |(1-e(an/q))+e(an/q)(1-e(\beta n))| \ge |1-e(an/q)| \Big(1-\frac{|1-e(n\beta)|}{|1-e(an/q)|}\Big)= 2\Big|\sin\frac{\pi an}{q}\Big| \Big( 1- \frac{|\sin(\pi n\beta)|}{|\sin(\pi an/q)|}\Big). $$ Now write $\Vert x\Vert= \min_{\ell \in {\Bbb Z}} |x-\ell|$ for the distance from $x$ to the nearest integer. Note also that for $0\le x\le \pi/2$ one has $2x/\pi \le \sin x \le x$. Therefore we get for $1\le n\le N$ $$ |1-e(n\alpha)| \ge |1-e(an/q)| \Big(1 - \frac{\pi n|\beta|}{2\Vert an/q\Vert}\Big) \ge |1-e(an/q)| \exp\Big( - \frac{\pi q|\beta|}{\Vert an/q\Vert}\Big), \tag{3} $$ where in the last inequality we used that $\eta =\pi n|\beta|/(2\Vert an/q\Vert) \le \pi q^2|\beta|/2 \le 1/10$, so that $(1-\eta) \ge \exp(-2\eta)$.

Multiplying (3) for $n$ from $1$ to $N=q-1$, we obtain $$ \prod_{n=1}^{N} |1-e(n\alpha)| \ge \prod_{n=1}^{q-1} |1-e(an/q)| \exp\Big( - \sum_{n=1}^{N} \frac{\pi q |\beta|}{\Vert an/q\Vert}\Big). \tag{4} $$ Now $$ \sum_{n=1}^{q-1} \frac{1}{\Vert an/q\Vert} \le 2\sum_{n\le q/2} \frac{q}{n} =2 q \log q +O(q), $$ and $\prod_{n=1}^{N} |1-e(an/q)| = q$, and so from (4) we conclude that $$ \prod_{n=1}^{q-1} |1-e(n\alpha)| \ge q \exp\Big( - 2\pi q^2 |\beta| \log q +O(q^2|\beta|)\Big) \gg q^{0.9}. $$

This proves the claim. Note that this argument is closely related to the nice attempt of Sangchul Lee.

The product does not tend to the limit zero. For any irrational number $\alpha$ one can show that $$ \limsup_{N\to \infty} \prod_{n=1}^{N} |1- e(n\alpha)| = \infty. \tag{1} $$ (Here I use the usual notation $e(\alpha) =e^{2\pi i\alpha}$, and the product in the question is stated for $\alpha/2$ rather than $\alpha$).

I'll prove a slightly weaker result; namely I'll show that (1) holds for any irrational $\alpha$ for which there are infinitely many rational approximations $a/q$ with $(a,q)=1$ and $$ \alpha=\frac{a}{q} +\beta, \qquad \text{with}\qquad |\beta| \le \frac{1}{100q^2}. \tag{2} $$ The $1/100$ is just chosen for ease of exposition, and a more careful argument can make do with $1/\sqrt{5}$ (any constant $<1/2$ is enough), and then it is well known that every irrational number admits infinitely many approximations with $|\alpha -a/q| \le 1/(\sqrt{5}q^2)$. Condition 2 holds for almost all irrational numbers -- it fails only if the continued fraction expansion only uses numbers below $100$.

Suppose then that $q$ is such that (2) holds, and consider the product in question at $N=q-1$. By the triangle inequality, for $1\le n\le N$, $$ |1-e(n\alpha)| = |(1-e(an/q))+e(an/q)(1-e(\beta n))| \ge |1-e(an/q)| \Big(1-\frac{|1-e(n\beta)|}{|1-e(an/q)|}\Big)= 2\Big|\sin\frac{\pi an}{q}\Big| \Big( 1- \frac{|\sin(\pi n\beta)|}{|\sin(\pi an/q)|}\Big). $$ Now write $\Vert x\Vert= \min_{\ell \in {\Bbb Z}} |x-\ell|$ for the distance from $x$ to the nearest integer. Note also that for $0\le x\le \pi/2$ one has $2x/\pi \le \sin x \le x$. Therefore we get for $1\le n\le N$ $$ |1-e(n\alpha)| \ge |1-e(an/q)| \Big(1 - \frac{\pi n|\beta|}{2\Vert an/q\Vert}\Big) \ge |1-e(an/q)| \exp\Big( - \frac{\pi q|\beta|}{\Vert an/q\Vert}\Big), \tag{3} $$ where in the last inequality we used that $\eta =\pi n|\beta|/(2\Vert an/q\Vert) \le \pi q^2|\beta|/2 \le 1/10$, so that $(1-\eta) \ge \exp(-2\eta)$.

Multiplying (3) for $n$ from $1$ to $N=q-1$, we obtain $$ \prod_{n=1}^{N} |1-e(n\alpha)| \ge \prod_{n=1}^{q-1} |1-e(an/q)| \exp\Big( - \sum_{n=1}^{N} \frac{\pi q |\beta|}{\Vert an/q\Vert}\Big). \tag{4} $$ Now $$ \sum_{n=1}^{q-1} \frac{1}{\Vert an/q\Vert} \le 2\sum_{n\le q/2} \frac{q}{n} =2 q \log q +O(q), $$ and $\prod_{n=1}^{N} |1-e(an/q)| = q$, and so from (4) we conclude that $$ \prod_{n=1}^{q-1} |1-e(n\alpha)| \ge q \exp\Big( - 2\pi q^2 |\beta| \log q +O(q^2|\beta|)\Big) \gg q^{0.9}. $$

This proves the claim. Note that this argument is closely related to the nice attempt of Sangchul Lee.

Added information (October, 2021): From comments on Fedja's recent question Can all partial sums $\sum_{k=1}^n f(ka)$ where $f(x)=\log|2\sin(x/2)|$ be non-negative? I was led to the following paper by D.S. Lubinsky which shows that (see Theorem 1.2 there) $$ \limsup_{N\to \infty} \frac{\log \prod_{n=1}^{N} |1-e(n\alpha)|}{\log N} \ge 1, $$ for all irrational numbers $\alpha$, so that the product in the question gets almost as large as $N$ infinitely often.

The product does not tend to the limit zero. For any irrational number $\alpha$ one can show that $$ \limsup_{N\to \infty} \prod_{n=1}^{N} |1- e(n\alpha)| = \infty. \tag{1} $$ (Here I use the usual notation $e(\alpha) =e^{2\pi i\alpha}$, and the product in the question is stated for $\alpha/2$ rather than $\alpha$).

I'll prove a slightly weaker result; namely I'll show that (1) holds for any irrational $\alpha$ for which there are infinitely many rational approximations $a/q$ with $(a,q)=1$ and $$ \alpha=\frac{a}{q} +\beta, \qquad \text{with}\qquad |\beta| \le \frac{1}{100q^2}. \tag{2} $$ The $1/100$ is just chosen for ease of exposition, and a more careful argument can make do with $1/\sqrt{5}$, and then it is well known that every irrational number admits infinitely many approximations with $|\alpha -a/q| \le 1/(\sqrt{5}q^2)$. Condition 2 holds for almost all irrational numbers -- it fails only if the continued fraction expansion only uses numbers below $100$.

Suppose then that $q$ is such that (2) holds, and consider the product in question at $N=q-1$. By the triangle inequality, for $1\le n\le N$, $$ |1-e(n\alpha)| = |(1-e(an/q))+e(an/q)(1-e(\beta n))| \ge |1-e(an/q)| \Big(1-\frac{|1-e(n\beta)|}{|1-e(an/q)|}\Big)= 2\Big|\sin\frac{\pi an}{q}\Big| \Big( 1- \frac{|\sin(\pi n\beta)|}{|\sin(\pi an/q)|}\Big). $$ Now write $\Vert x\Vert= \min_{\ell in {\Bbb Z}} |x-\ell|$ for the distance from $x$ to the nearest integer. Note also that for $0\le x\le \pi/2$ one has $2x/\pi \le \sin x \le x$. Therefore we get for $1\le n\le N$ $$ |1-e(n\alpha)| \ge |1-e(an/q)| \Big(1 - \frac{\pi n|\beta|}{2\Vert an/q\Vert}\Big) \ge |1-e(an/q)| \exp\Big( - \frac{\pi q|\beta|}{\Vert an/q\Vert}\Big), \tag{3} $$ where in the last inequality we used that $\eta =\pi n|\beta|/(2\Vert an/q\Vert) \le \pi q^2|\beta|/2 \le 1/10$, so that $(1-\eta) \ge \exp(-2\eta)$.

Multiplying (3) for $n$ from $1$ to $N=q-1$, we obtain $$ \prod_{n=1}^{N} |1-e(n\alpha)| \ge \prod_{n=1}^{q-1} |1-e(an/q)| \exp\Big( - \sum_{n=1}^{N} \frac{\pi q |\beta|}{\Vert an/q\Vert}\Big). \tag{4} $$ Now $$ \sum_{n=1}^{q-1} \frac{1}{\Vert an/q\Vert} \le 2\sum_{n\le q/2} \frac{q}{n} =2 q \log q +O(q), $$ and $\prod_{n=1}^{N} |1-e(an/q)| = q$, and so from (4) we conclude that $$ \prod_{n=1}^{q-1} |1-e(n\alpha)| \ge q \exp\Big( - 2\pi q^2 |\beta| \log q +O(q^2|\beta|)\Big) \gg q^{0.9}. $$

This proves the claim. Note that this argument is closely related to the nice attempt of Sangchul Lee.

The product does not tend to the limit zero. For any irrational number $\alpha$ one can show that $$ \limsup_{N\to \infty} \prod_{n=1}^{N} |1- e(n\alpha)| = \infty. \tag{1} $$ (Here I use the usual notation $e(\alpha) =e^{2\pi i\alpha}$, and the product in the question is stated for $\alpha/2$ rather than $\alpha$).

I'll prove a slightly weaker result; namely I'll show that (1) holds for any irrational $\alpha$ for which there are infinitely many rational approximations $a/q$ with $(a,q)=1$ and $$ \alpha=\frac{a}{q} +\beta, \qquad \text{with}\qquad |\beta| \le \frac{1}{100q^2}. \tag{2} $$ The $1/100$ is just chosen for ease of exposition, and a more careful argument can make do with $1/\sqrt{5}$ (any constant $<1/2$ is enough), and then it is well known that every irrational number admits infinitely many approximations with $|\alpha -a/q| \le 1/(\sqrt{5}q^2)$. Condition 2 holds for almost all irrational numbers -- it fails only if the continued fraction expansion only uses numbers below $100$.

Suppose then that $q$ is such that (2) holds, and consider the product in question at $N=q-1$. By the triangle inequality, for $1\le n\le N$, $$ |1-e(n\alpha)| = |(1-e(an/q))+e(an/q)(1-e(\beta n))| \ge |1-e(an/q)| \Big(1-\frac{|1-e(n\beta)|}{|1-e(an/q)|}\Big)= 2\Big|\sin\frac{\pi an}{q}\Big| \Big( 1- \frac{|\sin(\pi n\beta)|}{|\sin(\pi an/q)|}\Big). $$ Now write $\Vert x\Vert= \min_{\ell \in {\Bbb Z}} |x-\ell|$ for the distance from $x$ to the nearest integer. Note also that for $0\le x\le \pi/2$ one has $2x/\pi \le \sin x \le x$. Therefore we get for $1\le n\le N$ $$ |1-e(n\alpha)| \ge |1-e(an/q)| \Big(1 - \frac{\pi n|\beta|}{2\Vert an/q\Vert}\Big) \ge |1-e(an/q)| \exp\Big( - \frac{\pi q|\beta|}{\Vert an/q\Vert}\Big), \tag{3} $$ where in the last inequality we used that $\eta =\pi n|\beta|/(2\Vert an/q\Vert) \le \pi q^2|\beta|/2 \le 1/10$, so that $(1-\eta) \ge \exp(-2\eta)$.

Multiplying (3) for $n$ from $1$ to $N=q-1$, we obtain $$ \prod_{n=1}^{N} |1-e(n\alpha)| \ge \prod_{n=1}^{q-1} |1-e(an/q)| \exp\Big( - \sum_{n=1}^{N} \frac{\pi q |\beta|}{\Vert an/q\Vert}\Big). \tag{4} $$ Now $$ \sum_{n=1}^{q-1} \frac{1}{\Vert an/q\Vert} \le 2\sum_{n\le q/2} \frac{q}{n} =2 q \log q +O(q), $$ and $\prod_{n=1}^{N} |1-e(an/q)| = q$, and so from (4) we conclude that $$ \prod_{n=1}^{q-1} |1-e(n\alpha)| \ge q \exp\Big( - 2\pi q^2 |\beta| \log q +O(q^2|\beta|)\Big) \gg q^{0.9}. $$

This proves the claim. Note that this argument is closely related to the nice attempt of Sangchul Lee.