Approach 1
Evaluate $\sum_{\ell=1}^\infty \frac{1}{\ell(e^{2\pi \ell}-1)}$
The above link gave me the answer for a modified series, but if I were to adapt this to our case then it reduces to $\displaystyle\sum_{\ell\ge1}\frac1{\ell(e^{\ell\ln 2}-1)}$ and there's no imaginary number $i$ involved, so it becomes very a different case. So that didn't lead me anywhere.
Approach 2
I tried the geometric series approach by replacing $\frac1{2^\ell-1}$ with $\sum_{k\ge1}2^{-k\ell}$. But then it becomes the double summation:
$$\sum_{\ell\ge0}\sum_{k\ge1}\frac{2^{-k\ell}}\ell.$$
Upon swapping the summation we get the summation
$$\sum_{k\ge1}\ln(1-2^{-k}).$$
which is where I started to get to the summation in the question title above.
Show $\ln2 = \sum_\limits{n=1}^\infty\frac1{n2^n}$
The above link has the question closest to mine, for which I could find an existing answer underneath but I go around in circles or arrive at the above sum-of-logs.
Approach 3
Comparing to the appropriate integral leads to
$$\int_1^\infty\frac{dx}{x(2^x-1)}.$$
Substitution didn't do the trick either (on trying $u=2^x$ as well as $u=2^x-1$).
Approach 4
Is there function $f$ which has a Taylor expansion about $0$ with a radius of convergence move than 1, such that $\displaystyle f^{(\ell)}(0)=\frac{(\ell-1)!}{2^\ell-1}$?
If yes, that would do the trick, because then the summation is simply $f(1)$.
Approach 5
A good reference for an end-to-end computation would be nice. I couldn't find that either.
I am answering the question in the comments. It has to be shown that $$ - \sum\limits_{k = 1}^\infty {\ln (1 - 2^{ - k} )} < \ln 4. $$ Note that for $0<x<\frac{1}{2}$, $$ - \ln (1 - x) = x + \frac{{x^2 }}{2} + x^3 \int_0^1 \!{\frac{{t^2 }}{{1 - xt}}dt} \le x + \frac{{x^2 }}{2} + x^3 \int_0^1\! {\frac{{t^2 }}{{1 - t/2}}dt} < x + \frac{{x^2 }}{2} + \frac{{2x^3 }}{3}. $$ Then $$ - \sum\limits_{k = 1}^\infty {\ln (1 - 2^{ - k} )} < \sum\limits_{k = 1}^\infty {2^{ - k} } + \frac{1}{2}\sum\limits_{k = 1}^\infty {4^{ - k} } + \frac{2}{3}\sum\limits_{k = 1}^\infty {8^{ - k} } = \frac{{53}}{{42}} < 1.3 < \ln 4. $$