It's fairly easy to evaluate, to do this expand the logs in Taylor series in $x=(k+1)^{-2}$:
$$ \log_2(1-x) = \frac{-1}{\log 2}\sum_{m\geq1}\frac{x^{m}}{m}$$
$$ \log_2(-\log_2(1-x)) = \frac{\log x}{\log 2} - \frac{\log\log 2}{\log 2} + \sum_{n\geq 1}a_n x^n, $$
where $a_n$ are Taylor series coefficients of the l.h.s. after the log-singularity is subtracted. These can be easily calculated directly using numerical quadrature (code below).
Using the identities (here, as above, $x=(k+1)^{-2}$)
$$ \sum_{k\geq 1}x^s = \zeta(2s)-1, $$
$$ \sum_{k\geq 1}x^s\log x = 2\zeta'(2s), $$
we can rewrite the goal sum as a sum of three terms:
$$ \frac{-1}{(\log 2)^2}\sum_{k,m\geq 1}\frac{x^m\log x}{m} = \frac{-1}{(\log 2)^2} \sum_{m\geq 1}\frac{2\zeta'(2s)}{m} = 4.067782509260337209451548799089911685899767925643908, $$
the second
$$ \frac{\log\log 2}{(\log 2)^2}\sum_{k,m\geq 1}\frac{x^m}{m} = \frac{\log\log 2}{\log 2} = -0.52876637294489761424749777977881481518723706368332, $$
and a triple sum with separately calculated coefficients:
$$ \frac{-1}{\log 2}\sum_{k,m,n\geq1} \frac{a_n}{m}x^{m+n} = \frac{-1}{\log 2}\sum_{m,n\geq1}\frac{a_n}{m}(\zeta(2m+2n)-1) = -0.1064886215397004957702703848332558106111776391548. $$
Because $\zeta(a)-1 = O(2^{-a})$, these sums converge extremely quickly, and can be evaluated directly without numerical extrapolation techniques, yielding the result
$$ 3.4325275147757390994337806344778410601013532228057895 $$
It looks like it matches the number your link gives ($3.432527514776$) to at least the digits given there. I used fifty digits, and the most time-consuming portion of this is calculating the derivatives.
from mpmath import mp
from mpmath import *
import mpmath
mp.dps = 50
lg = lambda x: mp.log(x, b=2)
@mpmath.memoize
def A(n):
if n == 0:
return 0
return (mp.diff(lambda x: lg(-lg(1-x))-(log(x)-log(log(2)))/log(2), 0, n, method="quad")/mp.gamma(n+1)).real
s1 = -log(2)**(-2)*nsum(lambda m: 2*zeta(2*m, derivative=1)/m, (1, mp.inf))
s2 = log(log(2))/log(2)**2 * nsum(lambda m: (zeta(2*m)-1)/m, (1, mp.inf))
s3 = -1/log(2) * nsum(monitor(lambda m, n: A(n)/m * (zeta(2*m+2*n)-1)), (1, mp.inf), (1, mp.inf), method="direct")
print(s1 + s2 + s3)