On calculating the limit of the infinite product $\prod_{k=3}^n (1-\tan^4\frac{\pi}{2^k})$

You obtained an approximation of the exact value. In order to find such exact vale, note that $$(1-\tan^4(\alpha/2))=(1+\tan^2(\alpha/2))(1-\tan^2(\alpha/2))= \frac{4}{\cos(\alpha)}\left(\frac{\tan(\alpha/2)}{\tan(\alpha)}\right)^2.$$ Hence, as $n$ goes to infinity, $$\prod_{k=3}^n \left(1-\tan^4(\pi/2^k)\right)=\frac{4^{n-2}}{\prod_{k=3}^n\cos(\pi/2^{k-1})}\cdot \left(\prod_{k=3}^n\frac{\tan(\pi/2^k)}{\tan(\pi/2^{k-1})}\right)^2\\=4^{n-2}\cdot 2^{n-2}\sin(\pi/2^{n-1})\cdot \left(\frac{\tan(\pi/2^n)}{\tan(\pi/2^{2})}\right)^2\to \frac{\pi^3}{32}$$ where we used the known fact that $$\prod\limits_{k=2}^{n}\cos\left(\frac{\pi }{2^{k}}\right)= \frac{1}{2^{n-1}\sin(\pi/2^n)}$$ (see for example How to evaluate $\lim\limits_{n\to \infty}\prod\limits_{r=2}^{n}\cos\left(\frac{\pi}{2^{r}}\right)$).