Following Mike Spivey's comment above I will consider $B(n)$ in (3). It turns out that your conjecture is true, because for odd $n$ the sum $B(n)$ is a certain weighted $L^2$ norm of the Eulerian polynomial of the second kind of order $\frac{n+1}{2},$ with the sign of $(-1)^{\frac{n-1}{2}}\ . $
The product of the two factorials in $B(n)$ may be expressed in terms of the Eulerian Beta integral $$(m+1)!(2n−m)!=(2n+2)!\int_0^1 t^{m+1 }(1-t)^{2n-m}dt,$$ so that dividing it by $(2n+2)! $ we have $$ \frac {B(n)}{(2n+2)!}=\ \int_0^1 \sum_{m\ge0}(-1)^m\left\langle\!\!\!\left\langle n\atop m\right\rangle\!\!\!\right\rangle\ t^{m+1 }(1-t)^{2n-m}dt = $$ $$= \int_0^1 t(1-t)^{2n}\sum_{m\ge0}\ \left\langle\!\!\!\left\langle n\atop m\right\rangle \!\!\!\right\rangle \ \Big(\frac{t}{t-1}\Big)^{\! m} dt= \int_0^1 t(1-t)^{2n} E_n \Big(\frac{t}{t-1}\Big) dt. $$ Changing variable with $x:=\frac{t}{t-1}$ this becomes: $$ \int_{-\infty}^0 x(x-1)^{-2n-3} E_n(x) dx, $$
where $E_n$ denotes the Eulerian polynomial of the second kind $$E_n(x):=\sum_{m\ge0}\ \left\langle\!\!\!\left\langle n\atop m\right\rangle \!\!\!\right\rangle \ x^m,$$ and satisfies the recursive relation (corresponding to the relation for the coefficient that you gave in your question): $$(x-1)^{-2n-2}E_{n+1}(x)=\left( -x(x-1)^{-2n-1}E_n(x) \right)^{\prime}.$$ By the above formula it is now easy to show, integrating by parts repeatedly, that $B(n)=0$ for even $n$ while for odd $n=2p-1$ $$B(2p-1)=(-1) ^ { p + 1 } (4p)! \int_0^{+\infty} E_p (-x)^2 (x+1)^{-4p-1} x \ dx\ . $$
(To check this, it is convenient to introduce the sequence of rational functions $U_ n(x):= (x-1)^{-2n}E_n(x)$ that satisfy the recurrence $U_ {n+1}= \big (\frac{x}{1-x} U_ n \big) ^ {\prime} $ with initial condition $U_0=1.$ Hence for all $n+m>0$ we have $\int_{-\infty}^0 U_{n+1}U_{m}\frac{x}{1-x}dx=-\int_{-\infty}^0 U_ {n}U_ {m+1}\frac{x}{1-x}dx\ .$ The integral found above for $B(n) / (2n+2)!\ $ was $\int_{-\infty}^0 -\int_{-\infty}^0 U_{n}U_{1}\frac{x}{1-x}dx$ ).
