I'll start by saying that I don't have an answer, but I found this interesting and wanted to share some of the manipulations I noticed. I'm essentially "thinking out loud". Perhaps in recasting the question, somebody else will notice something. The following is a mixture of manipulations made by hand and some simplifications done in Mathematica.
I started by Loging both sides to write this as a summation identity instead, and doing some simplification
$$
f(x) = 2^{\frac{1}{2^x}} \prod _{j=1}^{\infty } \prod _{k=0}^{2^{j-1}-1}
\left(1-\frac{1}{\left(2^j+2 k+1\right)^2}\right)^{(2^{x})^{-j-1}}
$$
$$
\ln( f(x) ) = \frac{1}{2^x} \ln(2)+\sum _{j=1}^{\infty } \sum _{k=0}^{2^{j-1}-1}
(2^{x})^{-j-1} \ln\left(1-\frac{1}{\left(2^j+2 k+1\right)^2}\right)
$$
$$
\ln( f(x) ) = \frac{1}{2^x} \ln(2)+\sum _{j=1}^{\infty } (2^{x})^{-j-1} \sum _{k=0}^{2^{j-1}-1}
\ln\left(1-\frac{1}{\left(2^j+2 k+1\right)^2}\right)
$$
The inner sum can now be evaluated symbolically in Mathematica:
Sum[Log[1 - 1/(2^j + 2 k + 1)^2], {k, 0, 2^(j - 1) - 1}]
Out[] = (1/2)*Log[2*Pi] - Log[Gamma[2^(-1 + j)]] + \
Derivative[1, 0][Zeta][0, 2^j] - \
Derivative[1, 0][Zeta][0, 1 + 2^(-1 + j)] - \
2*Derivative[1, 0][Zeta][0, 1/2 + 2^j] + \
2*Derivative[1, 0][Zeta][0, (1/2)*(1 + 2^j)] + \
Derivative[1, 0][Zeta][0, 1 + 2^j]
Mathematica won't simplify these derivatives, but we can, using
$$
\frac{\partial}{\partial s} \zeta(0,a) = \ln(\Gamma(a))-\frac{1}{2}\ln(2\pi)
$$
The factors of $\frac{1}{2}\ln(2\pi)$ all cancel, leaving
$$
\ln\Gamma(2^j)-\ln\Gamma(2^{j-1})+\ln\Gamma(1+2^j)-\ln\Gamma(1+2^{j-1})-2(\ln\Gamma(\frac{1}{2}+2^j)-\ln\Gamma(\frac{1}{2}+2^{j-1}))
$$
and now Mathematica can help us out:
FullSimplify[
Log[Gamma[2^j]/Gamma[2^(j - 1)]] +
Log[Gamma[1 + 2^j]/Gamma[1 + 2^(j - 1)]] -
2 Log[Gamma[1/2 + 2^j]/Gamma[1/2 + 2^(j - 1)]], {j >= 1}]
Out[] = Log[(2^(-1 + 2^(1 + j))*Gamma[(1/2)*(1 + 2^j)]^4)/(Pi*
Gamma[1/2 + 2^j]^2)]
Which we can rewrite
$$
\log \left(\frac{2^{2^{j+1}-1} \Gamma \left(\frac{1}{2} \left(1+2^j\right)\right)^4}{\pi \Gamma
\left(\frac{1}{2}+2^j\right)^2}\right) = \left(2^{j+1}-1\right) \log (2)+2 \log \left(\frac{\Gamma \left(\frac{1}{2}+2^{j-1}\right)^2}{\Gamma
\left(\frac{1}{2}+2^j\right)}\right)-\log (\pi )
$$
Our summation is now:
$$
\ln( f(x) ) = \frac{1}{2^x} \ln(2)+\sum _{j=1}^{\infty } (2^{x})^{-j-1} \left(
\left(2^{j+1}-1\right) \log (2)+2 \log \left(\frac{\Gamma \left(\frac{1}{2}+2^{j-1}\right)^2}{\Gamma
\left(\frac{1}{2}+2^j\right)}\right)-\log (\pi ) \right)
$$
and the simpler sums can be evaluated and simplified in Mathematica:
$$
\sum _{j=1}^{\infty } \left(2^{j+1}-1\right) \log (2) \left(2^x\right)^{-j-1} = \frac{2^{-x} \left(3\ 2^x-2\right) \log (2)}{-3\ 2^x+4^x+2}
$$
$$
\sum _{j=1}^{\infty } \log (\pi ) \left(2^x\right)^{-j-1} = \frac{\log (\pi )}{4^x-2^x}
$$
leaving
$$
\ln( f(x) ) = \frac{1}{2^x} \ln(2) + \frac{2^{-x} \left(3\ 2^x-2\right) \log (2)}{-3\ 2^x+4^x+2} - \frac{\log (\pi )}{4^x-2^x} + 2 \sum _{j=1}^{\infty } (2^{x})^{-j-1}
\log \left(\frac{\Gamma \left(\frac{1}{2}+2^{j-1}\right)^2}{\Gamma
\left(\frac{1}{2}+2^j\right)} \right)
$$
or:
$$
\ln( f(x) ) = \frac{\left(2^{1-x}-1\right) \log (\pi )+2^x \log (2)}{-3\ 2^x+4^x+2} + 2 \sum _{j=1}^{\infty } (2^{x})^{-j-1}
\log \left(\frac{\Gamma \left(\frac{1}{2}+2^{j-1}\right)^2}{\Gamma
\left(\frac{1}{2}+2^j\right)} \right)
$$
At this point, I started playing around with manipulations of the $\Gamma$ functions, and had some success using identities for the half-integer values to rewrite in terms of factorials, but I noticed that my math must have gone awry somewhere, as the evaluated term has discontinuities at $x=0$ and $x=1$ and the summation seems to also converge at $x=0$. I think this is coming close to a closed form of $f(x)$ but something is slightly off.
EDIT:
A bit of further experimentation revealed that the equation
$$
\log \left(\frac{2^{2^{j+1}-1} \Gamma \left(\frac{1}{2} \left(1+2^j\right)\right)^4}{\pi \Gamma
\left(\frac{1}{2}+2^j\right)^2}\right)
$$
is where my trouble began. This form can be evaluated numerically as follows:
f[x_] := Exp[Log[2]/2^x + NSum[(2^x)^(-j - 1) Log[(2^(2^(j + 1) - 1) \
Gamma[1/2 + 2^(j - 1)]^4)/(Pi Gamma[1/2 + 2^j]^2)], {j, 1, Infinity}]]
{f[0],f[1]}
(* {1.5708, 1.35914} *)
but if you try to split the terms up, discontinuities start to creep in. Even just pulling out the $\ln(\pi)$ term "breaks" the evaluation at $x=0$:
NSum[(2^0)^(-j -
1) Log[(2^(2^(j + 1) - 1) Gamma[1/2 + 2^(j - 1)]^4)/(Gamma[
1/2 + 2^j]^2)], {j, 1, Infinity}]
(*ComplexInfinity*)
If a closed form exists, it will likely involve the limiting behaviour of several discontinuous functions.
Gamma[]andZeta[]. – Corey Kelly Jun 12 '13 at 11:28