I doubt there exists a closed formula for $n\ne 2$. In the case $n=2$ such formula exists only thanks to the double-angle formula for cosine.

Let $n$ be fixed and $c=c_n$. Notice that $n=c^2-c$ and $c\to\infty$ as soon as $n\to\infty$.

Denote by $p_k$ the $k$-th multiplier in the product $S_n$. It can be easily seen that $$c\cdot (p_k^2-1) = p_{k-1} - 1$$

Let $$f(x) = 1 + x + \frac{x^2}{2(2c-1)} + \frac{x^3}{2(2c-1)^2(2c+1)} + \frac{x^4(2c+5)}{8(2c-1)^3(2c+1)(4c^2+2c+1)} + \dots$$ be a solution to the functional equation: $$c\cdot(f(x)^2-1)=f(2cx)-1$$ with $f(0)=1$ and $f'(0)=1$. Then $$p_k = f\left(\frac{x_0}{(2c)^k}\right)$$ where $x_0$ is a solution to $f(x_0)=0$.

Now $$S_n = \prod_{k=1}^{\infty} p_k = \exp \sum_{k=1}^{\infty} \ln\left(1 + \Theta\left(\frac{x_0}{(2c)^k}\right) \right) = \exp \sum_{k=1}^{\infty} \Theta\left(\frac{x_0}{(2c)^k}\right) = \exp \Theta\left(\frac{x_0}{2c-1}\right)$$ which tends to $1$ as $c\to\infty$.

Therefore, $S_n\to 1$ as $n\to\infty$.

I doubt there exists a closed formula for $n\ne 2$. In the case $n=2$ such formula exists only thanks to the double-angle formula for cosine.

Let $n$ be fixed and $c=c_n$. Notice that $n=c^2-c$ and $c\to\infty$ as soon as $n\to\infty$.

Denote by $p_k$ the $k$-th multiplier in the product $S_n$. It can be easily seen that $$c\cdot (p_k^2-1) = p_{k-1} - 1$$

Consider the functional equation: $$c\cdot(f(x)^2-1)=f(2cx)-1$$ with $f(0)=1$ and $f'(0)=1$. Its solution can be expressed as a series: $$f(x) = 1 + x + \frac{x^2}{2(2c-1)} + \frac{x^3}{2(2c-1)^2(2c+1)} + \frac{x^4(2c+5)}{8(2c-1)^3(2c+1)(4c^2+2c+1)} + \dots.$$ Then $$p_k = f\left(\frac{x_0}{(2c)^k}\right)$$ where $x_0$ is a solution to $f(x_0)=0$.

Now $$S_n = \prod_{k=1}^{\infty} p_k = \exp \sum_{k=1}^{\infty} \ln\left(1 + \Theta\left(\frac{x_0}{(2c)^k}\right) \right) = \exp \sum_{k=1}^{\infty} \Theta\left(\frac{x_0}{(2c)^k}\right) = \exp \Theta\left(\frac{x_0}{2c-1}\right)$$ which tends to $1$ as $c\to\infty$.

Therefore, $S_n\to 1$ as $n\to\infty$.

Example. For $n=2$, the functional equation admits the analytic solution $f(x)=\cosh(\sqrt{2x})$ for which $x_0=\frac{-\pi^2}{8}$.