Using the integral representation of Bernoulli numbers I obtain formally the integral representation of the double summation
$$
\sum_{k=1}^{\infty}\sum_{j=0}^{k}\binom{k}{j}\frac{B_{j+k+1}}{j+k+1}=2\cdot\int_0^\infty\frac{t}{e^{2 \pi  t}-1}\frac{dt}{t^2+(1+t^2)^2}=0.069591059035995961110566767049...
$$
So the alternative form of the question is
$$
\int_0^\infty\frac{t}{e^{2 \pi  t}-1}\frac{dt}{t^2+(1+t^2)^2}=\frac14+\frac{\ln\phi}{1-2\phi}
$$
$\it{Proof}.$ Since
$$
\frac{1}{t^2+(1+t^2)^2}=\frac{1}{\sqrt{5}}\left(\frac1{t^2+1/\phi^2}-\frac1{t^2+\phi^2}\right),
$$
and according to Binet's second integral representation for the digamma function $\psi$
$$
\psi(\phi)=-\frac1{2\phi}+\ln\phi-2\int_0^\infty\frac{t}{e^{2 \pi  t}-1}\frac{dt}{t^2+\phi^2},
$$
$$
\psi(1/\phi)=-\frac{\phi}{2}-\ln\phi-2\int_0^\infty\frac{t}{e^{2 \pi  t}-1}\frac{dt}{t^2+1/\phi^2},
$$
and 
$$
\psi(\phi)-\psi(1/\phi)=\frac1{\phi-1}
$$
one has 
\begin{align}
&\int_0^\infty\frac{t}{e^{2 \pi  t}-1}\frac{dt}{t^2+(1+t^2)^2}\\
&=\frac1{2\sqrt5}\left(\psi(\phi)-\psi(1/\phi)+\frac1{2\phi}-\frac{\phi}2-2\ln\phi\right)\\
&=\frac14+\frac{\ln\phi}{1-2\phi}.
\end{align}