To have here an alternative approach, I'll cross-post my answer from MSE here, adapted for the series. Similarly to what @Nemo obtained in the comments, by utilizing the identities:

$$\sqrt 3 \frac{\arcsin \left(\frac{\sqrt 3}{2} x\right)}{\sqrt{1-\frac{3}{4} x^2}} = \sum_{n=1}^\infty \frac{3^n x^{2n-1}}{n\binom{2n}{n}}$$

$$\int_0^\frac{\pi}{2}\left(4\sin^{4n-3}x-2\sin^{4n-1}x\right)dx=\frac{16^n}{(2n-1)\binom{4n}{2n}}$$

We can transform the original series into an equivalent integral. Furthermore, in order to evaluate the resulting integral we will use a long chain of substitutions, those come with the goal in mind to arrive at a point where we can smoothly switch to double integrals by using:

$$ \frac{\arctan\left(f(x)\frac{\sqrt{g(x)}}{\sqrt{h(x)}} \right)}{\sqrt{g(x)}\sqrt{h(x)}}dx=\int_0^{f(x)} \frac{1}{h(x)+g(x)y^2}dy$$

$$\sum_{n=1}^\infty \frac{48^n }{n(2n-1)\binom{2n}{n}\binom{4n}{2n}}=2\sqrt 3 \int_0^\frac{\pi}{2} \frac{2-\sin^2 x}{\sin x}\frac{\arcsin \left(\frac{\sqrt 3}{2} \sin^2 x\right)}{\sqrt{1-\frac{3}{4} \sin^4x}}dx$$

$$\overset{\sin^2 x\to \cos x}=\sqrt 6\int_0^\frac{\pi}{2}\frac{(2-\cos x)\cos \frac{x}{2}}{\cos x}\frac{\arcsin \left(\frac{\sqrt 3}{2} \cos x\right)}{\sqrt{1-\frac{3}{4} \cos^2x}}dx$$

$$\overset{\frac{\pi}{2}-x\to \arcsin x}=\sqrt 6\int_0^1 \frac{\color{blue}{(1-x)^{3/2}}+(1+x)^{3/2}}{(1-x^2)\sqrt{1+3x^2}}\arcsin\left(\frac{\sqrt 3\sqrt{1-x^2}}{2}\right)dx$$

$$\overset{{\color{blue}{x\to -x}}}=\sqrt 6\int_{-1}^1 \frac{(1+x)^{3/2} \arcsin\left(\small \frac{\sqrt 3\sqrt{1-x^2}}{2}\right)}{(1-x^2)\sqrt{1+3x^2}}dx \overset{x\to \frac{1-x}{1+x}} = \sqrt 3\int_0^\infty \frac{\arcsin\left(\frac{\sqrt{3x}}{1+x}\right)}{x\sqrt{1+x^3}}dx$$

$$=\sqrt{3}\int_0^\infty \frac{\arctan \left(\frac{\color{red}{\sqrt{3x(1+x)}}}{\sqrt{1+x^3}}\right)}{x\sqrt{1+x^3}}dx=\sqrt{3}\int_0^\infty \int_0^{\color{red}{f(x)}} \frac{1}{x(1+x^3+y^2)}dydx$$

$$\overset{y\to f(y)}=\sqrt{3} \int_0^\infty \int_0^x \frac{f'(y)}{x(1+x^3+f^2(y))}dydx=\int_0^\infty \int_0^x (*)dydx = \int_0^\infty \int_y^\infty (*)dxdy$$

$$ = \frac{1}{\sqrt 3}\int_0^\infty \left(\frac{\ln \left(\frac{x^3}{1+x^3+f^2(y)}\right)}{1+f^2(y)}\right) \bigg|_{x\to y}^{x\to \infty} dy = \frac{3}{2}\int_0^\infty \frac{(1+2y)\ln\left(\frac{1+y}{y}\right)}{\sqrt{y(1+y)}(1+3y+3y^2)}dy$$

$$\overset{\frac{y}{1+y}\to y^2}=-6 \int_0^1 \frac{(1+y^2)\ln y}{1+y^2+y^4}dy =2\sqrt{3}\sum_{n=1}^\infty \frac{\sin\left(\frac{n\pi}{3}\right)+\sin\left(\frac{2n\pi}{3}\right)}{n^2}$$

$$= 2\sqrt{3}\left(\operatorname{Cl}_2\left(\frac{\pi}{3}\right)+\operatorname{Cl}_2\left(\frac{2\pi}{3}\right)\right)=\frac{10}{\sqrt 3}\operatorname{Cl}_2\left(\frac{\pi}{3}\right) = \boxed{\frac{15}{2} L(2,\chi)}$$

Where $\operatorname{Cl}_2(x)$ is the Clausen Function.

To provide an alternative approach, I'll cross-post my answer from MSE, adapted for the series.
Similarly to what @Nemo obtained in the comments, we can transform the original series into an equivalent integral, by utilizing the following identities:

$$\sqrt 3 \frac{\arcsin \left(\frac{\sqrt 3}{2} x\right)}{\sqrt{1-\frac{3}{4} x^2}} = \sum_{n=1}^\infty \frac{3^n x^{2n-1}}{n\binom{2n}{n}}$$

$$\int_0^\frac{\pi}{2}\left(4\sin^{4n-3}x-2\sin^{4n-1}x\right)dx=\frac{16^n}{(2n-1)\binom{4n}{2n}}$$

Furthermore, to evaluate the resulting integral we will use a long chain of substitutions aimed to arrive at a point where we can smoothly switch to double integrals by using:

$$ \frac{\arctan\left(f(x)\frac{\sqrt{g(x)}}{\sqrt{h(x)}} \right)}{\sqrt{g(x)}\sqrt{h(x)}}dx=\int_0^{f(x)} \frac{1}{h(x)+g(x)y^2}dy$$

$$\sum_{n=1}^\infty \frac{48^n }{n(2n-1)\binom{2n}{n}\binom{4n}{2n}}=2\sqrt 3 \int_0^\frac{\pi}{2} \frac{2-\sin^2 x}{\sin x}\frac{\arcsin \left(\frac{\sqrt 3}{2} \sin^2 x\right)}{\sqrt{1-\frac{3}{4} \sin^4x}}dx$$

$$\overset{\sin^2 x\to \cos x}=\sqrt 6\int_0^\frac{\pi}{2}\frac{(2-\cos x)\cos \frac{x}{2}}{\cos x}\frac{\arcsin \left(\frac{\sqrt 3}{2} \cos x\right)}{\sqrt{1-\frac{3}{4} \cos^2x}}dx$$

$$\overset{\frac{\pi}{2}-x\to \arcsin x}=\sqrt 6\int_0^1 \frac{\color{blue}{(1-x)^{3/2}}+(1+x)^{3/2}}{(1-x^2)\sqrt{1+3x^2}}\arcsin\left(\frac{\sqrt 3\sqrt{1-x^2}}{2}\right)dx$$

$$\overset{{\color{blue}{x\to -x}}}=\sqrt 6\int_{-1}^1 \frac{(1+x)^{3/2} \arcsin\left(\small \frac{\sqrt 3\sqrt{1-x^2}}{2}\right)}{(1-x^2)\sqrt{1+3x^2}}dx \overset{x\to \frac{1-x}{1+x}} = \sqrt 3\int_0^\infty \frac{\arcsin\left(\frac{\sqrt{3x}}{1+x}\right)}{x\sqrt{1+x^3}}dx$$

$$=\sqrt{3}\int_0^\infty \frac{\arctan \left(\frac{\color{red}{\sqrt{3x(1+x)}}}{\sqrt{1+x^3}}\right)}{x\sqrt{1+x^3}}dx=\sqrt{3}\int_0^\infty \int_0^{\color{red}{f(x)}} \frac{1}{x(1+x^3+y^2)}dydx$$

$$\overset{y\to f(y)}=\sqrt{3} \int_0^\infty \int_0^x \frac{f'(y)}{x(1+x^3+f^2(y))}dydx=\int_0^\infty \int_0^x (*)dydx = \int_0^\infty \int_y^\infty (*)dxdy$$

$$ = \frac{1}{\sqrt 3}\int_0^\infty \left(\frac{\ln \left(\frac{x^3}{1+x^3+f^2(y)}\right)}{1+f^2(y)}\right) \bigg|_{x\to y}^{x\to \infty} dy = \frac{3}{2}\int_0^\infty \frac{(1+2y)\ln\left(\frac{1+y}{y}\right)}{\sqrt{y(1+y)}(1+3y+3y^2)}dy$$

$$\overset{\frac{y}{1+y}\to y^2}=-6 \int_0^1 \frac{(1+y^2)\ln y}{1+y^2+y^4}dy =2\sqrt{3}\sum_{n=1}^\infty \frac{\sin\left(\frac{n\pi}{3}\right)+\sin\left(\frac{2n\pi}{3}\right)}{n^2}$$

$$= 2\sqrt{3}\left(\operatorname{Cl}_2\left(\frac{\pi}{3}\right)+\operatorname{Cl}_2\left(\frac{2\pi}{3}\right)\right)=\frac{10}{\sqrt 3}\operatorname{Cl}_2\left(\frac{\pi}{3}\right) = \boxed{\frac{15}{2} L(2,\chi)}$$

Where $\operatorname{Cl}_2(x)$ is the Clausen Function and $L(s,\chi)$ is the Dirichlet L-function.

To have here an alternative approach, I'll cross-post my answer from MSE here. Similarly to what @Nemo obtained in the comments, by utilizing the identities:

$$\sqrt 3 \frac{\arcsin \left(\frac{\sqrt 3}{2} x\right)}{\sqrt{1-\frac{3}{4} x^2}} = \sum_{n=1}^\infty \frac{3^n x^{2n-1}}{n\binom{2n}{n}}$$

$$\int_0^\frac{\pi}{2}\left(4\sin^{4n-3}x-2\sin^{4n-1}x\right)dx=\frac{16^n}{(2n-1)\binom{4n}{2n}}$$

We can transform the original series into an equivalent integral. Furthermore, in order to evaluate the resulting integral we will use a long chain of substitutions, those come with the goal in mind to arrive at a point where we can switch to double integrals by using:

$$ \frac{\arctan\left(f(x)\frac{\sqrt{g(x)}}{\sqrt{h(x)}} \right)}{\sqrt{g(x)}\sqrt{h(x)}}dx=\int_0^{f(x)} \frac{1}{h(x)+g(x)y^2}dy$$

$$\sum_{n=1}^\infty \frac{48^n }{n(2n-1)\binom{2n}{n}\binom{4n}{2n}}=2\sqrt 3 \int_0^\frac{\pi}{2} \frac{2-\sin^2 x}{\sin x}\frac{\arcsin \left(\frac{\sqrt 3}{2} \sin^2 x\right)}{\sqrt{1-\frac{3}{4} \sin^4x}}dx$$

$$\overset{\sin^2 x\to \cos x}=\sqrt 6\int_0^\frac{\pi}{2}\frac{(2-\cos x)\cos \frac{x}{2}}{\cos x}\frac{\arcsin \left(\frac{\sqrt 3}{2} \cos x\right)}{\sqrt{1-\frac{3}{4} \cos^2x}}dx$$

$$\overset{\frac{\pi}{2}-x\to \arcsin x}=\sqrt 6\int_0^1 \frac{\color{blue}{(1-x)^{3/2}}+(1+x)^{3/2}}{(1-x^2)\sqrt{1+3x^2}}\arcsin\left(\frac{\sqrt 3\sqrt{1-x^2}}{2}\right)dx$$

$$\overset{{\color{blue}{x\to -x}}}=\sqrt 6\int_{-1}^1 \frac{(1+x)^{3/2} \arcsin\left(\small \frac{\sqrt 3\sqrt{1-x^2}}{2}\right)}{(1-x^2)\sqrt{1+3x^2}}dx \overset{x\to \frac{1-x}{1+x}} = \sqrt 3\int_0^\infty \frac{\arcsin\left(\frac{\sqrt{3x}}{1+x}\right)}{x\sqrt{1+x^3}}dx$$

$$=\sqrt{3}\int_0^\infty \frac{\arctan \left(\frac{\color{red}{\sqrt{3x(1+x)}}}{\sqrt{1+x^3}}\right)}{x\sqrt{1+x^3}}dx=\sqrt{3}\int_0^\infty \int_0^{\color{red}{f(x)}} \frac{1}{x(1+x^3+y^2)}dydx$$

$$\overset{y\to f(y)}=\sqrt{3} \int_0^\infty \int_0^x \frac{f'(y)}{x(1+x^3+f^2(y))}dydx=\int_0^\infty \int_0^x (*)dydx = \int_0^\infty \int_y^\infty (*)dxdy$$

$$ = \frac{1}{\sqrt 3}\int_0^\infty \left(\frac{\ln \left(\frac{x^3}{1+x^3+f^2(y)}\right)}{1+f^2(y)}\right) \bigg|_{x\to y}^{x\to \infty} dy = \frac{3}{2}\int_0^\infty \frac{(1+2y)\ln\left(\frac{1+y}{y}\right)}{\sqrt{y(1+y)}(1+3y+3y^2)}dy$$

$$\overset{\frac{y}{1+y}\to y^2}=-6 \int_0^1 \frac{(1+y^2)\ln y}{1+y^2+y^4}dy =2\sqrt{3}\sum_{n=1}^\infty \frac{\sin\left(\frac{n\pi}{3}\right)+\sin\left(\frac{2n\pi}{3}\right)}{n^2}$$

$$= 2\sqrt{3}\left(\operatorname{Cl}_2\left(\frac{\pi}{3}\right)+\operatorname{Cl}_2\left(\frac{2\pi}{3}\right)\right)=\frac{10}{\sqrt 3}\operatorname{Cl}_2\left(\frac{\pi}{3}\right) = \boxed{\frac{15}{2} L(2,\chi)}$$

Where $\operatorname{Cl}_2(x)$ is the Clausen Function.

To have here an alternative approach, I'll cross-post my answer from MSE here, adapted for the series. Similarly to what @Nemo obtained in the comments, by utilizing the identities:

$$\sqrt 3 \frac{\arcsin \left(\frac{\sqrt 3}{2} x\right)}{\sqrt{1-\frac{3}{4} x^2}} = \sum_{n=1}^\infty \frac{3^n x^{2n-1}}{n\binom{2n}{n}}$$

$$\int_0^\frac{\pi}{2}\left(4\sin^{4n-3}x-2\sin^{4n-1}x\right)dx=\frac{16^n}{(2n-1)\binom{4n}{2n}}$$

We can transform the original series into an equivalent integral. Furthermore, in order to evaluate the resulting integral we will use a long chain of substitutions, those come with the goal in mind to arrive at a point where we can smoothly switch to double integrals by using:

$$ \frac{\arctan\left(f(x)\frac{\sqrt{g(x)}}{\sqrt{h(x)}} \right)}{\sqrt{g(x)}\sqrt{h(x)}}dx=\int_0^{f(x)} \frac{1}{h(x)+g(x)y^2}dy$$

$$\sum_{n=1}^\infty \frac{48^n }{n(2n-1)\binom{2n}{n}\binom{4n}{2n}}=2\sqrt 3 \int_0^\frac{\pi}{2} \frac{2-\sin^2 x}{\sin x}\frac{\arcsin \left(\frac{\sqrt 3}{2} \sin^2 x\right)}{\sqrt{1-\frac{3}{4} \sin^4x}}dx$$

$$\overset{\sin^2 x\to \cos x}=\sqrt 6\int_0^\frac{\pi}{2}\frac{(2-\cos x)\cos \frac{x}{2}}{\cos x}\frac{\arcsin \left(\frac{\sqrt 3}{2} \cos x\right)}{\sqrt{1-\frac{3}{4} \cos^2x}}dx$$

$$\overset{\frac{\pi}{2}-x\to \arcsin x}=\sqrt 6\int_0^1 \frac{\color{blue}{(1-x)^{3/2}}+(1+x)^{3/2}}{(1-x^2)\sqrt{1+3x^2}}\arcsin\left(\frac{\sqrt 3\sqrt{1-x^2}}{2}\right)dx$$

$$\overset{{\color{blue}{x\to -x}}}=\sqrt 6\int_{-1}^1 \frac{(1+x)^{3/2} \arcsin\left(\small \frac{\sqrt 3\sqrt{1-x^2}}{2}\right)}{(1-x^2)\sqrt{1+3x^2}}dx \overset{x\to \frac{1-x}{1+x}} = \sqrt 3\int_0^\infty \frac{\arcsin\left(\frac{\sqrt{3x}}{1+x}\right)}{x\sqrt{1+x^3}}dx$$

$$=\sqrt{3}\int_0^\infty \frac{\arctan \left(\frac{\color{red}{\sqrt{3x(1+x)}}}{\sqrt{1+x^3}}\right)}{x\sqrt{1+x^3}}dx=\sqrt{3}\int_0^\infty \int_0^{\color{red}{f(x)}} \frac{1}{x(1+x^3+y^2)}dydx$$

$$\overset{y\to f(y)}=\sqrt{3} \int_0^\infty \int_0^x \frac{f'(y)}{x(1+x^3+f^2(y))}dydx=\int_0^\infty \int_0^x (*)dydx = \int_0^\infty \int_y^\infty (*)dxdy$$

$$ = \frac{1}{\sqrt 3}\int_0^\infty \left(\frac{\ln \left(\frac{x^3}{1+x^3+f^2(y)}\right)}{1+f^2(y)}\right) \bigg|_{x\to y}^{x\to \infty} dy = \frac{3}{2}\int_0^\infty \frac{(1+2y)\ln\left(\frac{1+y}{y}\right)}{\sqrt{y(1+y)}(1+3y+3y^2)}dy$$

$$\overset{\frac{y}{1+y}\to y^2}=-6 \int_0^1 \frac{(1+y^2)\ln y}{1+y^2+y^4}dy =2\sqrt{3}\sum_{n=1}^\infty \frac{\sin\left(\frac{n\pi}{3}\right)+\sin\left(\frac{2n\pi}{3}\right)}{n^2}$$

$$= 2\sqrt{3}\left(\operatorname{Cl}_2\left(\frac{\pi}{3}\right)+\operatorname{Cl}_2\left(\frac{2\pi}{3}\right)\right)=\frac{10}{\sqrt 3}\operatorname{Cl}_2\left(\frac{\pi}{3}\right) = \boxed{\frac{15}{2} L(2,\chi)}$$

Where $\operatorname{Cl}_2(x)$ is the Clausen Function.