First of all I do not understand why you say the product converges to 1. In fact the product must me strictly less than the first term namely, $(1-\tan^4(\frac{\pi}{8}))$. You can however say this function converges since $\displaystyle \sum_{k=1} \tan^4(\frac{\pi}{2^k}) < \infty$. Hence the product converges absolutely.
(Note: $\displaystyle \Pi_{k=1}^n (1+a_k) \leq \exp(\displaystyle \sum_{k=1}^n a_k)$. So convergence of $\displaystyle \sum_{k=1}^{\infty} a_k$ implies the convergence of the infinite product $\displaystyle \Pi_{k=1}^{\infty} (1+a_k)$)
I tried to find an approximate answer using Matlab and then make an educated guess about the limit and then try to prove it. However, the limit of the first $10^7$ terms give me $0.968946146259369$.
I am unable to make an educated guess (something based on $\pi$,$e$ or the golden ratio). It will be interesting to see what this product converges to!
$\textbf{EDIT:}$
This is just for the sake of completion and the solution is due to $\textbf{Moron}$. Following the idea suggested by Moron, and since we know the product has to converge let
$p_n = \displaystyle \Pi_{k=1}^n (1-\tan^4(\frac{\pi}{2^{k+2}})) = \displaystyle \Pi_{k=1}^n \frac{\cos(\frac{\pi}{2^{k+1}})}{\cos^4(\frac{\pi}{2^{k+2}})} = \frac{\cos(\frac{\pi}{4})}{\cos(\frac{\pi}{2^{n+2}})} \frac{1}{\displaystyle \Pi_{k=1}^n \cos^3(\frac{\pi}{2^{k+2}})}$.
$p_n = \frac{\cos(\frac{\pi}{4})}{\cos(\frac{\pi}{2^{n+2}})} \frac{8^n \times \sin^3(\frac{\pi}{2^{n+2}})}{\sin^3(\frac{\pi}{4})}$
Take the limit as $n \rightarrow \infty$.
Hence, $p = \frac{\cos(\frac{\pi}{4})}{1} \times \frac{(\frac{\pi}{4})^3}{\sin^3(\frac{\pi}{4})}$.
Hence, $p = 2 \times (\frac{\pi}{4})^3 = \frac{\pi^3}{32} \approx 0.968946146259369$