Given that, $$\sum_{j=0}^{\infty}\frac{2^j\left(j-\frac{1}{3}\right)^3\left(j^2+j-1\right)}{(2j+1)(2j+3){2j \choose j}}=A\tag1$$
We have $A=2\pi+12+\frac{1}{3}?$
We can generalize the above $(1)$:
$$\sum_{j=0}^{\infty}\frac{2^jj^n}{(2j+1)(2j+3){2j \choose j}}=F(n)\tag2$$
I believe the closed for $(2)$ is
$$F(n)=(-1)^{n}(2n+1)-(-1)^n (2n-1)\cdot \frac{\pi}{2}$$
How can we show that the proposed $(2)$ is correct?
On
Too long for a comment.
Just out of curiosity, I computed $$G_n=2(-1)^n\sum_{j=0}^{\infty}\frac{2^jj^n}{(2j+1)(2j+3){2j \choose j}}$$ and got the surprising results $$\left( \begin{array}{cc} 0 & 4-\pi \\ 1 & 6-2 \pi \\ 2 & 10-3 \pi \\ 3 & 14-5 \pi \\ 4 & 26-6 \pi \\ 5 & 14-17 \pi \\ 6 & 186+27 \pi \\ 7 & -1042-380 \pi \\ 8 & 10906+3399 \pi \\ 9 & -119218-38057 \pi \\ 10 & 1490042+474132 \pi \\ 11 & -20706898-6591455 \pi \\ 12 & 317327706+101008179 \pi \\ 13 & -5315948530-1692119522 \pi \\ 14 & 96656651578+30766766937 \pi \\ 15 & -1895852523154-603468602105 \pi \\ 16 & 39903246182426+12701597748834 \pi \\ 17 & -897111926876722-285559595341037 \pi \\ 18 & 21456828988070394+6829920793053567 \pi \\ 19 & -544015833214835410-173165617952800580 \pi \\ 20 & 14574701046924559066+4639271431409321739 \pi \end{array} \right)$$ where, as already said in comments and answers, no pattern seems to appear.
From the expansion of $\arcsin^2 t$ \begin{equation} \arcsin^2 t=\sum_{p=0}^\infty \frac{2^{2p}t^{2p+2}}{(2p+1)(p+1)\binom{2p}{p}} \end{equation} we can obtain by differentiation \begin{equation} 2\frac{\arcsin t}{\sqrt{1-t^2}}=\sum_{p=0}^\infty \frac{2^{2p+1}t^{2p+1}}{(2p+1)\binom{2p}{p}} \end{equation} Multiplying the above identity by $t$ and integrating, one obtains \begin{equation} 2\int_0^x\frac{t\arcsin t}{\sqrt{1-t^2}}\,dt=\sum_{p=0}^\infty \frac{2^{2p+1}x^{2p+3}}{(2p+1)(2p+3)\binom{2p}{p}} \end{equation} We choose $x=\sqrt{y}/2$ and divide the identity by $y^{3/2}$ to write \begin{equation} \frac{8}{y^{3/2}}\int_0^{\sqrt{y}/2}\frac{t\arcsin t}{\sqrt{1-t^2}}\,dt=\sum_{p=0}^\infty \frac{y^{p}}{(2p+1)(2p+3)\binom{2p}{p}} \end{equation} Now, applying the operator $y\frac{d}{dy}$ $n$ times and taking the result at $y=2$ gives \begin{equation} \left.\left[ y\frac{d}{dy}\right]^n\frac{8}{y^{3/2}}\int_0^{\sqrt{y}/2}\frac{t\arcsin t}{\sqrt{1-t^2}}\,dt\right|_{y=2}=\sum_{p=0}^\infty \frac{p^{n}2^p}{(2p+1)(2p+3)\binom{2p}{p}} \end{equation} The function can be evaluated as \begin{equation} \frac{8}{y^{3/2}}\int_0^{\sqrt{y}/2}\frac{t\arcsin t}{\sqrt{1-t^2}}\,dt=\frac{4}{y}-8\frac{\sqrt{1-\frac{y}{4}}}{y^{3/2}}\arcsin\left( \frac{\sqrt{y}}{2} \right) \end{equation} The above result can be simplified by taking $z=\sqrt{y}/2$: \begin{equation} \left.\left[\frac{1}{2} z\frac{d}{dz}\right]^n \left[ \frac{1}{z^2}-\frac{\sqrt{1-z^2}}{z^3}\arcsin z\right] \right|_{z=\sqrt{2}/2}=\sum_{p=0}^\infty \frac{p^{n}2^p}{(2p+1)(2p+3)\binom{2p}{p}} \end{equation} As $\arcsin \left(\sqrt{2}/2\right)=\pi/4$ and since the successive applications of the operator on $z^{-2}$ and $z^{-3}\left( 1-z^2 \right)^k$ ($k$ is an integer) at $z=1/\sqrt{2}$ give rational results, we expect \begin{equation} F(n)=a_n+b_n\pi \end{equation} where $a_n$ and $b_n$ are rational. I could not find any simple expression for these coefficients however. First few values are $F(n)=2-\pi/2,-3+\pi,5-3\pi/2,-7+5\pi/2,13-3\pi,-7+17\pi/2, 93+27\pi/2\cdots$ for $n=0,1,2,3,4,5,6\cdots$. They do not correspond to the propose formula as remarked by @Mariusz Iwaniuk in a comment.
With $z=\exp(t/2)$, we have $1/2zd/dz=d/dt$, and we can build the generating function for the $F(n)$: \begin{equation} \sum_{n=0}^\infty F(n)\frac{t^n}{n!}=\phi\left( t-\ln2 \right) \end{equation} where \begin{equation} \phi\left( t-\ln 2 \right)=2e^{-3t/2}\left[ e^{t/2}-\sqrt{2-e^{t}}\arcsin\left(\frac{ e^{t/2}}{\sqrt{2}} \right)\right] \end{equation}