Xrfgtjtk

Given that, $$sum_{j=0}^{infty}frac{2^jleft(j-frac{1}{3}right)^3left(j^2+j-1right)}{(2j+1)(2j+3){2j choose j}}=Atag1$$

We have $A=2pi+12+frac{1}{3}?$

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?

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}
2frac{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}
2int_0^xfrac{tarcsin 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{tarcsin 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 $yfrac{d}{dy}$ $n$ times and taking the result at $y=2$ gives
begin{equation}
left.left[ yfrac{d}{dy}right]^nfrac{8}{y^{3/2}}int_0^{sqrt{y}/2}frac{tarcsin t}{sqrt{1-t^2}},dtright|_{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{tarcsin t}{sqrt{1-t^2}},dt=frac{4}{y}-8frac{sqrt{1-frac{y}{4}}}{y^{3/2}}arcsinleft( 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} zfrac{d}{dz}right]^n
left[ frac{1}{z^2}-frac{sqrt{1-z^2}}{z^3}arcsin zright]
right|_{z=sqrt{2}/2}=sum_{p=0}^infty frac{p^{n}2^p}{(2p+1)(2p+3)binom{2p}{p}}
end{equation}
As $arcsin left(sqrt{2}/2right)=pi/4$ 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_npi
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 $F(n)=2-pi/2,-3+pi,5-3pi/2,-7+5pi/2,13-3pi,-7+17pi/2, 93+27pi/2cdots$ for $n=0,1,2,3,4,5,6cdots$. 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!}=phileft( t-ln2 right)
end{equation}
where
begin{equation}
phileft( t-ln 2 right)=2e^{-3t/2}left[ e^{t/2}-sqrt{2-e^{t}}arcsinleft(frac{ e^{t/2}}{sqrt{2}} right)right]
end{equation}

Too long for a comment.

Just out of curiosity, I computed $$G_n=2(-1)^nsum_{j=0}^{infty}frac{2^jj^n}{(2j+1)(2j+3){2j choose j}}$$ 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.