Finding a sum of $1+\frac{1}{4\cdot2^{4}}+\frac{1}{7\cdot2^{7}}+\frac{1}{10\cdot2^{10}}+\cdots$

To apply a discrete Fourier transform to Taylor series of $-\log(1-x)$ is a good idea.
Since $$ -\log(1-x)=\sum_{n\geq 1}\frac{x^n}{n} $$ we have $$ \sum_{n\geq 0}\frac{x^{3n+1}}{3n+1} = \int_{0}^{x}\frac{dt}{1-t^3}\\= -\frac{1}{3}\log(1-x)+\frac{1}{18} \left(-\sqrt{3} \pi +6 \sqrt{3} \arctan\left(\frac{1+2 x}{\sqrt{3}}\right)+3 \log\left(1+x+x^2\right)\right) $$ and by evaluating at $x=\frac{1}{2}$ it follows that $$ \sum_{n\geq 0}\frac{1}{(3n+1)2^{3n+1}}=\frac{1}{2}\,\phantom{}_2 F_1\left(\frac{1}{3},1;\frac{4}{3};\frac{1}{8}\right)= \color{red}{\frac{1}{18} \left[-\pi\sqrt{3}+6 \sqrt{3} \arcsin\left(\frac{2}{\sqrt 7}\right)+3 \log(7)\right]}$$

Similar to the answer by Jack D'Aurizio and yours, but let $$f(x) = \sum_{n=0}^\infty \frac{1}{3n+1}\cdot x^{3n+1}\implies f'(x) = \sum_{n=0}^\infty x^{3n} =\frac{1}{1-x^3}$$ Hence, $$f(x) = \int\frac{1}{1-x^3}dx = \frac{1}{6}\left(\ln(x^2+x+1)-2\ln(1-x)+2\sqrt{3}\arctan\left(\frac{2x+1}{\sqrt{3}}\right)\right)+C$$ Because $f(0)=0$ we have $C = -\frac{\sqrt{3}}{18}\pi$. We get our result by noticing that our result is $\frac{1}{2}+f\left(\frac{1}{2}\right)$, which is approximately $1.016849...$.

Sometimes when faced by one of these "sections" of a well-known series it helps to think "roots of unity". The other solutions given here are surely the way to go in this case, but it's worth knowing the generic technique, so here goes.

Let $G(x)=\sum_0^\infty \frac{x^{n+1}}{n+1}$ we see that we want "one-third" of this series. So with $\omega:=\exp{\frac{2\pi}{3}}$ we have that the given series is the value at $x=\frac{1}{2}$ of $$\frac{1}{3}\left[G(x)+\omega^2\ G(\omega x)+ \omega\ G(\omega^2 x)\right].$$ We'd get a couple of other series by taking the other obvious multipliers, $(1,1,1)$ and $(1,\omega,\omega^2)$.

As $G(x)=\log (1-x)$ there's then nothing left to do apart from some tedious arithmetic evaluating the modulus and argument of things like $1-\frac{1}{2}\omega$.