All the logarithms

$\begin{aligned} f(x) & = \sum_{n=1}^\infty \ln \left(\frac 1{\sqrt[n2^n]{|x|}}\right) \\ & = \sum_{n=1}^\infty - \frac {\ln |x|}{n2^n} & \small \color{#3D99F6} \text{By Maclaurin series} \\ & = \ln \left(1-\frac 12\right)\ln|x| \\ & = - \ln 2 \ln|x| \end{aligned}$

The area bounded by $f(x)$ and the $x$ -axis is given by:

$\begin{aligned} I & = \int_{-1}^1 f(x) \ dx \\ & = - \ln 2 \int_{-1}^1 \ln |x| \ dx & \small \color{#3D99F6} \text{Since the integrand is even} \\ & = - 2 \ln 2 \int_0^1 \ln |x| \ dx \\ & = - 2 \ln 2 (x\ln x - x)\bigg|_0^1 \\ & = - 2 \ln 2 \left(1\ln 1 - 1 - {\color{#3D99F6}\lim_{x \to 0} x\ln x} + 0\right) & \small \color{#3D99F6} \text{By L'Hôpital's rule (see note)} \\ & = - 2 \ln 2 \left(0 - 1 - {\color{#3D99F6}0} + 0\right) \\ & = 2\ln 2 \\ & = \ln 4 \end{aligned}$

$\implies A = \boxed{4}$

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Note:

$\begin{aligned} \lim_{x \to 0} x\ln x & = \lim_{x \to 0} \frac {\ln x}{\frac 1x} & \small \color{#3D99F6} \text{A }\infty/\infty \text{ case, L'Hôpital's rule applies.} \\ & = \lim_{x \to 0} \frac {\frac 1x}{-\frac 1{x^2}} & \small \color{#3D99F6} \text{Differentiate up and down w.r.t. }x \\ & = \lim_{x \to 0} - \frac 1x\times x^2 \\ & = 0 \end{aligned}$