I have no idea how to approach the general problem, but here is a quick observation:

**A.** Let $\alpha = 2\beta$ so that $\beta$ is irrational if and only if $\alpha$ is so. Define $f$ by $f(x) = \log|2\sin\pi x|$. Then

$$ \log \left| \prod_{k=1}^{n} (1 - e^{\pi k i \alpha} ) \right| = \sum_{k=1}^{n} f(k\beta). $$

Now by the *Riemann-integrable criterion for equidistributed sequences*, for any $R$ we know that

$$ \varlimsup_{n\to\infty} \frac{1}{n} \sum_{k=1}^{n} f(k\beta)
\leq \lim_{n\to\infty} \frac{1}{n} \sum_{k=1}^{n} \max\{ f(k\beta), -R \}
= \int_{0}^{1} \max\{f(x), -R\} \, dx. $$

Taking $R \to \infty$, we get

$$ \varlimsup_{n\to\infty} \frac{1}{n} \sum_{k=1}^{n} f(k\beta) \leq \int_{0}^{1} f(x) \, dx = 0. \tag{1} $$

In view of (1), if the sequence $(k\beta \text{ mod } 1 : k \geq 1)$ is very evenly distributed over $\Bbb{R}/\Bbb{Z}$, then we even expect equality in (1). For me, it seems to suggest that we need some hard analysis on it.

**B**. Let $\beta$ be as before, and write $ n\beta = p_n + \epsilon_n$, where $p_n \in \Bbb{Z}$ and $|\epsilon_n| < 1/2$. Also, let us consider only $n$ which are *record-breaking* indices for $|\epsilon_n|$ in the sense that $|\epsilon_k| > |\epsilon_n|$ for all $k < n$.

We denote $a_n (\beta) = \left| \prod_{k=1}^{n} (1 - e^{2\pi i k \beta} ) \right|$ and then claim the following:

**Claim.** There exists constants $C_1, C_2 > 0$ such that for all irrational $\beta$ and for all record-breaking $n$,
$$ a_n(\beta) \leq C_1 |\epsilon_n| n \exp( C_2 |\epsilon_n| n \log n ). $$

In view of the *Dirichlet approximation theorem*, we know that $n|\epsilon_n| < 1$. This gives $a_n(\beta) \leq C_1 n^{C_2}$.

If (a subsequence of) $|\epsilon_n|$ decays at least as fast as $1/(n\log n)$ then we have $\liminf_n a_n = 0$. In particular, for any number $\beta$ with irrational measure greater than 2, this is true. Unfortunately, most interesting numbers are either proved or expected to have irrationality measure 2, so this says nothing anything about OP's question.

Now let $\tilde{\beta}_n = p_n / n$. For brevity, we remove the subscripts from $p_n$, $\epsilon_n$ and $\tilde{\beta}_n$ whenever no confusion arises. Then using the relation $|1 - e^{2ix}| = 2|\sin x|$ for any $x \in \Bbb{R}$, we can write

\begin{align*}
a_n(\beta)
&= 2|\sin (\pi n \beta)| \left| \prod_{k=1}^{n-1} (1 - e^{2\pi i k \tilde{\beta}} ) \right| \left| \prod_{k=1}^{n-1} \frac{\sin(\pi k \beta)}{\sin(\pi k \tilde{\beta})} \right| \\
&= 2n|\sin(\pi \epsilon)| \prod_{k=1}^{n-1} \left| \frac{\sin(\pi k \tilde{\beta} + \pi k \epsilon / n)}{\sin(\pi k \tilde{\beta})} \right| \\
&= 2n|\sin(\pi \epsilon)| \prod_{k=1}^{n-1} \left| \cos(\pi k \epsilon / n) + \cot(\pi k \tilde{\beta}) \sin(\pi k \epsilon / n) \right| \tag{2}
\end{align*}

In the second step, the following observation is used.

**Lemma 1.** If $n$ is record-breaking, then $n$ and $p_n$ are coprime.

*Proof.* Assume otherwise. Write $n = d\tilde{n}$ and $p_n = d\tilde{p}$ for $d = \gcd(n, p_n) > 1$. Then $\tilde{n}\beta = \tilde{p} + (\epsilon_n / d)$ and thus $|\epsilon_{\tilde{n}}| = |\epsilon_n| / d < |\epsilon_n|$, contradicting the assumption that $n$ is record-breaking. ////

Since $n$ and $p$ are coprime, we know that $\{ p, 2p, \cdots, (n-1)p \} \equiv \{1, 2, \cdots, n-1 \} \pmod n$. In particular, by permuting the indices we have

$$ \prod_{k=1}^{n-1} |1 - e^{2\pi i k p / n}|
= \prod_{k=1}^{n-1} |1 - e^{2\pi i k / n}|
= n. $$

We also notice that

$$\sum_{k=1}^{n} |\cot(\pi k / n)| = \mathcal{O}(n \log n). \tag{3} $$

This follows from the inequality $\cot x \leq 1/x$ on $(0, \pi/2]$. Combining these observations, we can bound the product term of (2). Indeed, it is clear that $|\sin(\pi k \epsilon / n)| \leq \pi |\epsilon|$ for any $1 \leq k \leq n$. Thus

\begin{align*}
\prod_{k=1}^{n-1} \left| \cos(\pi k \epsilon / n) + \cot(\pi k \tilde{\beta}) \sin(\pi k \epsilon / n) \right|
&\leq \prod_{k=1}^{n-1} ( 1 + \pi |\epsilon| |\cot(\pi k \tilde{\beta})| ) \\
&\leq \exp \left( \sum_{k=1}^{n-1} \pi |\epsilon| |\cot(\pi k / n)| \right) \\
&\leq e^{C|\epsilon|n \log n}.
\end{align*}

Here, we used Lemma 1 to rearrange the product and then used (3) to bound the exponent. This gives the desired upper bound in Claim.