Let :
$$B=\cdot\frac{1}{\sqrt{1+\frac{1}{\sqrt{1+\frac{1}{\sqrt{1+\frac{1}{\sqrt{1+\cdots}}}}}}}}$$
Let :
$$D=\frac{1}{B+\frac{1}{B+\frac{1}{B+\frac{1}{B+\cdots}}}}$$
Then I conjecture with WA that the limit is :
$$D=-\pi+\ln(16)+(7\ln\pi)/2-7/3\arctan(\pi)$$
Edit : Now after accepted answer there is a motivation the Newman's conjecture related to littlewood polynomials with coefficient $\pm 1$.
Question :
If my conjecture is true how to prove it ? If not does $D$ admits a closed form ? If yes what ?
If the continued fraction converges, then $B=\frac{1}{\sqrt{1+B}}$, so $B^3+B^2-1=0$. You can check this cubic has only one positive root. Thus, $B$ must be equal to that positive root, which, by Cardano's formula, is $$B=\frac{1}{6} \left(-2+\sqrt[3]{4(25-3 \sqrt{69})}+ \sqrt[3]{4(25+3 \sqrt{69})}\right).$$
Similarly, if the continued fraction converges for $D$, then $D=\frac{1}{B+D}$, so $D^2+BD-1=0$. The positive solution $D$ for this quadratic is $$D=\frac{-B+\sqrt{B^2+4}}{2}.$$ You can check that your conjectured limit is approximately equal to $0.6914204058\color{red}{720947}$. Unfortunately, the value of $D$ is $0.6914204058\color{red}{550671}$, making your conjecture false but quite impressively near to the correct value!
On the other hand, if you want a simplified form for $D$, you should try first to denest the cubic radicals appearing in $B$'s expression. I've tried to find some rationals $u$ and $v$ such that $(u+v\sqrt{69})^3=25-3\sqrt{69}$, but with no success. Mathematica's function for denesting radicals also doesn't give a simplified form, although this function doesn't always work.