We begin by using the identity $\tan(x) = \frac{\sin(x)}{\cos(x)}$. Substituting and remembering that $1=\frac{\cos^4(x)}{\cos^4(x)}$, the product becomes
$$\prod_{k=3}^{\infty} \frac{\cos^4 \left ( \frac{\pi}{2^k} \right ) -\sin^4 \left ( \frac{\pi}{2^k} \right ) }{\cos^4 \left ( \frac{\pi}{2^k} \right ) }.$$
Factoring, recognizing that $\cos^2(x)+\sin^2(x)= 1$, and recognizing that $\cos^2(x)-\sin^2(x) = \cos(2x)$, we obtain
$$\prod_{k=3}^{\infty} \frac{\cos \left ( \frac{\pi}{2^{k-1}} \right ) }{\cos^4 \left ( \frac{\pi}{2^k} \right ) }.$$ Since $\prod ab = \prod a \prod b$, we can split $\cos^3$ out and leave behind a telescoping product. Multiplying this out, we find that all terms cancel except for $\cos \left (\frac{\pi}{4} \right ) = \frac{\sqrt{2}}{2}$. Now, using the fact that $\prod a^n = \left (\prod a \right )^n$, we obtain
$$\frac{\sqrt{2}}{2} \left (\prod_{k=3}^{\infty} \cos \left (\frac{\pi}{2^k} \right ) \right )^{-3}.$$ Multiplying and dividing by 2 inside the cosine (and forgetting about $\frac{\sqrt{2}}{2}$ and the power of -3 for now), we have
$$\prod_{k=3}^{\infty} \cos \left ( \frac{\frac{\pi}{2}}{2^{k-1}} \right ).$$
Shifting the index by one, we get
$$\prod_{k=2}^{\infty} \cos \left ( \frac{\frac{\pi}{2}}{2^{k}} \right ).$$ The product is now almost in the form of
Viete’s formula,
$$ \prod_{k=1}^{\infty} \cos \left (\frac{\theta}{2^k} \right ) = \frac{\sin \theta}{\theta}.$$
However, our index starts from k=2, and not k=1, so we must first divide by $\cos \left ( \frac{\theta}{2} \right )$ to obtain
$$\prod_{k=2}^{\infty} \cos \left ( \frac{\theta}{2^k} \right ) = \frac{\sin \theta}{\theta \cos \left ( \frac{\theta}{2} \right )}.$$ Now, using the formula with $\theta = \frac{\pi}{2},$ we obtain
$$\frac{\sin \left ( \frac{\pi}{2} \right )}{\frac{\pi}{2} \cos \left ( \frac{\pi}{4} \right )} = \frac{4}{\pi \sqrt{2}}.$$
Putting it all together,
$$ \frac{\sqrt{2}}{2} \left ( \frac{4}{\pi \sqrt{2}} \right )^{-3} = \frac{\pi^3}{32}.$$