Expressions involving Eulerian numbers of the second kind: trying to show $\sum_{m=0}^{n} (-1)^m(m)m!(2n-m-2)!\left\langle\left\langle n\atop m\right\rangle\right\rangle\neq0$ for even $n$.
Surprise: your expression is a multiple of a Bernoulli number: $$A(n)=-\frac{(2n)!}{n}B_n $$ for all $n>1$, and the Bernoulli numbers are indeed vanishing exactly for all odd integers greater than 1. This follows computing the generating function for the $A(n)$, on the lines of the preceding answer and comments. Note that your identity can be written in the form $$\sum_{m=0}^n (-1)^n\frac{ \left\langle \!\!\left\langle n\atop m\right\rangle \!\!\right\rangle }{\left( 2n+1\atop m+1\right)}=2B_{n+1}\, .$$
After that, I had a vague memory of a similar relation about the Eulerian numbers of first kind, that it is the fourth identity reported here: $$\sum_{m=0}^n (-1)^n\frac{ \left\langle n\atop m\right\rangle }{\left( n\atop m\right)}=(n+1)B_n\, .$$
Checking the source of the latter may be useful for you, and will possibly give a reference for your identity (I'm quite confident that it should be written somewhere; hopefully some expert may recognize and provide a reference). In case, I will add the details of the (quite standard) computation of the generating function of yours $A(n)$ .
Details. Here is a generating function for yor sequence as a real function. This should allow a nice proof of the Bernoulli number formula, provided one is able to compute the last integral.
For $t\in\mathbb{R}$ and $x\in [0,\infty)$ let $$\psi(t,x):=W(x e^{x+t})=L^{-1}\big(t+L(x)\big)$$ where $W$ is the Lambert function and $L:\mathbb{R_+}\to\mathbb{R}$ is the invertible function $L(x):=x+\log(x)$. So $\psi$ is the general solution of the Cauchy problem for the autonomous ODE (the flow) $$\psi_t=\frac \psi {1+\psi} $$ $$\psi(0,x)=x$$ Thus it also solves the linear first-order PDE $$\psi_t- \frac x {1+x} \psi_x =0$$ and all derivatives w.r.to $t$ have the form $$\partial_t^n\psi= v_n(\psi)$$ for a recursively defined sequence $v_n(x)$ $$v_0(x)=x$$ $$v_{n+1}(x)=\frac x {1+x}v'_n(x)$$ So the $v_n$ are rational functions with poles at $-1$; $v_n(x)=O(x^{-n})$ as $x\to+\infty$ and in fact $$v_n(x)=-x(1+x)^{-2n+1}E_n(-x)$$ where $E_n$ are the Eulerian polynomial of second kind (these $v_n$ are just a simple modification of to the previously defined sequence $U_n$ ). For all $t\in\mathbb{R}$, all $r>0$ and $n\ge2$, it's easy to see that $$\sup_{|t|\le r}\, \big| v_n\big(\psi(t,x)\big) \big|:=g_{r,n}(x)\in L^1(\mathbb{R}_+),$$ which allows (by the dominated convergence theorem) to differentiate under the sign of integral the function $$h(t):=\int_0^\infty \psi_{tt}(t,x)dx=\int_0^\infty v_2\big(\psi(t,x)\big)dx\, ,$$ so that $$h^{(n)}(t)=\int_0^\infty v_{n+2}\big(\psi(t,x)\big)dx,$$ and in particular we have for $n\ge 1$ $$h^{n-1}(0)=\int_0^\infty v_{n+1}(x)dx=-\frac {A(n)}{(2n)!}$$
The relation $$-\frac {A(n)}{(2n)!} = \frac {B_n} n$$ for all $n\ge 1$ now writes:
$$\int_0^\infty \frac{W(xe^{x+t})}{\big(1+W(xe^{x+t})\big)^3}dx=\frac1 t - \frac 1 {e^t - 1}.$$
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 $-\int_{-\infty}^0 U_{n}U_{1}\frac{x}{1-x}dx$ ).