Following Martin Rubey's comment above I will compute $B(n)$ 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 you gave in your question for their coefficients:
$$(x-1)^{-2n-2}E_{n+1}(x)=\left( -x(x-1)^{-2n-1}E_n(x)  \right)^{\prime}.$$
By the above formula it is now easy to show, integrating by parts repeatedly, that $B(n)=0$ for even $n$ while for odd $n=2p-1$ 
$$B(2p-1)=(-1) ^ { p + 1 } (4p)! \int_0^{+\infty} E_p (-x)^2 (x+1)^{-4p-1} x \\ dx\\  . $$