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: $$f(x) = 1 + x + \frac{x^2}{2(2c-1)} + \frac{x^3}{2(2c-1)^2(2c+1)} + \frac{x^4(2c+5)}{8(2c-1)^3(2c+1)(4c^2+2c+1)} + \dots$$ be a solution to the functional equation: $$c\cdot(f(x)^2-1)=f(2cx)-1$$ with $f(0)=1$ and $f'(0)=1$. 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)$$which tends to 1 as c\to\infty. Therefore, S_n\to 1 as n\to\infty. Example. For n=2, the functional equation admits the analytic solution f(x)=\cosh(\sqrt{2x}) for which x_0=\frac{-\pi^2}{8}. 1 I doubt there exists a closed formula for n\ne 2. In the case n=2 such formula exists only thanks to the double-angle formula for cosine. Let n be fixed and c=c_n. Notice that n=c^2-c and c\to\infty as soon as n\to\infty. Denote by p_k the k-th multiplier in the product S_n. It can be easily seen that$$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 $n\to\infty$.