The closed form of the product is basically $$\prod_{k=1}^n\left(1-2z\cos\frac{(2k-1)\pi}{n}+z^2\right)=(1+z^n)^2,$$ shown using $(z-e^{i\phi})(z-e^{-i\phi})=1-2z\cos\phi+z^2$ and the roots of $z^n+1=0$.
Thus, to compute $f(\alpha)=\displaystyle\lim_{n\to\infty}\prod_{k=1}^n\big[2-(2-\alpha^2/n^2)\cos\big((2k-1)\pi/n\big)\big]$, we choose $z=z_n$ so that $1-2z\cos\phi+z^2$ is proportional to $2-(2-\alpha^2/n^2)\cos\phi$: $$\frac{2z_n}{1+z_n^2}=r_n:=1-\frac{\alpha^2}{2n^2}\implies z_n=\frac{1\pm\sqrt{1-r_n^2}}{r_n}.$$ Now we easily get $z_n=1\pm\alpha/n+o(1/n)$ as $n\to\infty$, hence $$f(\alpha)=\lim_{n\to\infty}\left(\frac2{1+z_n^2}\right)^n(1+z_n^n)^2=e^\alpha(1+e^{-\alpha})^2.$$
Your limit is $f\left(\sqrt{\pi^2-8}\right)\approx6.17966119933834352235124640584393+$.
This question is closely related to computing the discriminant of Chebyshev polynomials of the second kind, found in literature (e.g., see G. Szegő Orthogonal polynomials, theorem $6.71$).
I'm going an elementary way here. Denote $x_{n,k}=\cos(k\pi/n)$ for $0\leqslant k\leqslant n$, so that $$S_n(x)=2^{n-1}\prod_{k=0}^n(x-x_{n,k})$$ is our polynomial. Then "ignoring the $1$ in the $\sqrt{1+(\dots)^2}$" amounts to saying that $l_k:=l_{n,k}$ is approximately twice the supremum of $\big|S_n(x)\big|$ on $x_{n,k-1}<x<x_k$; more precisely, we have $$L=\lim_{n\to\infty}\prod_{k=1}^n l_{n,k}=\lim_{n\to\infty}\prod_{k=1}^n\color{LightGray}{\Big[}2\sup_{x_{n,k-1}<x<x_{n,k}}\big|S_n(x)\big|\color{LightGray}{\Big]}=\lim_{n\to\infty}2^n\prod_{k=1}^n \big|S_n(x_{n,k}')\big|,$$ where $x_{n,k}'$ (for $1\leqslant k\leqslant n$) are the roots of $S_n'(x)$.
From properties of resultants, we know that if $f$ is a polynomial of degree $d$, with leading coefficient $a$, roots $x_{d,k}$ for $1\leqslant k\leqslant d$, and roots of its derivative $x'_{d,k}$ for $0<k<d$, then $$\prod_{k=1}^{d-1}f(x_{d,k}')=\frac1{d^d a}\prod_{k=1}^d f'(x_{d,k}).$$
In our case $d=n+1$ and $a=2^{n-1}$, so that $$L=\lim_{n\to\infty}\frac{2L_n}{(n+1)^{n+1}},\qquad L_n=\prod_{k=0}^n\big|S_n'(x_{n,k})\big|.$$
From $S_n(\cos t)=-\sin t\sin nt$ we get $S_n'(\cos t)=n\cos nt+\cot t\sin nt$, hence $$S_n'(1)=2n,\quad S_n'(-1)=2n(-1)^n,\quad S_n'(x_{n,k})=n(-1)^k\quad(0<k<n)$$ and $L_n=4n^{n+1}$, thus $L=\lim\limits_{n\to\infty}8/(1+1/n)^{n+1}=8/e$ as claimed.
Best Answer
Let
$$K = \prod_{k=3}^\infty \frac1{\cos\left(\frac\pi k\right)}$$
As Bouwkamp points out under First method in this paper,
$$\cos(x) = \frac{\sin(2x)}{2x} \cdot \frac x{\sin(x)} \tag{1}$$
Then
$$\begin{align*} \frac1K &= \prod_{k=3}^\infty \cos\left(\frac\pi k\right) \\[1ex] &= \prod_{k=3}^\infty \frac{\sin\left(\frac{2\pi}k\right)}{\frac{2\pi}k} \cdot \prod_{k=3}^\infty \frac{\frac\pi k}{\sin\left(\frac\pi k\right)} \tag{1}\\[1ex] &= \left[\prod_{k=1}^\infty \frac{\sin\left(\frac{2\pi}{2k+1}\right)}{\frac{2\pi}{2k+1}} \cdot \prod_{k=2}^\infty \frac{\sin\left(\frac{2\pi}{2k}\right)}{\frac{2\pi}{2k}}\right] \cdot \prod_{k=3}^\infty \frac{\frac\pi k}{\sin\left(\frac\pi k\right)} \tag{2}\\[1ex] &= \left[\prod_{k=1}^\infty \frac{\sin\left(\frac{2\pi}{2k+1}\right)}{\frac{2\pi}{2k+1}}\right] \cdot \frac2\pi \end{align*}$$
where in $(2)$ we split the product over even and odd indices.