Define a sequence $\{x_i\}_i$ so that $x_1=1/2$ and
$$x_{n+1}=\frac{x_n^2+1}{2}.$$
We claim that
$$d_n=\frac{1-x_{n-1}}{x_{n-1}x_n}.$$
We show this by induction on $n$, with the base case $n=2$ true since $x_2=5/8$. For the inductive step, we show that, if $d=\frac{2(1-x)}{x(x^2+1)}$, then
$$\frac{}{}$$
is the largest root of the given polynomial. This is via the factorization of
$$2(5d^4+8d^2+8d+8)y^3+4d(5d^4+11d^3-2d^2+14d-4)y^2+4d^3(5d^3+2d^2+24d-16)y-32d^5(d+2)$$
with $d=\frac{2(1-x)}{x(x^2+1)}$ as
$$\frac{16}{x^6(x^2+1)^6}\big(y(x^2+1)(x^4+2x^2+5)-8(1-x^2)\big)\big(a(x)y^2+b(x)y+c(x)),$$
times $a(x)y^2+b(x)y+c(x)$, where
\begin{align*}
a(x)&=x^2 (x^2 + 1) (x^4 - 2 x^3 + 4 x^2 - 4 x + 2) (x^4 + 2 x^3 - 2 x + 1)\\
b(x)&=2 (x - 1)^2 x (x^7 + x^6 + 6 x^5 - 5 x^4 + 9 x^3 - 2 x^2 - 6 x + 4)\\
c(x)&=16(x-1)^4(x^2-x+1).
\end{align*}
It is not hard to see (for example, by graphing them) that $a(x),b(x),c(x)\geq 0$ as long as $x>0$; since $0<x_n<1$ for all $n$, this gives that
$$\frac{8(1-x^2)}{x^6+3x^4+7x^2+5}$$
is the largest root of the polynomial, so
$$d_{n+1}=\frac{8(1-x_{n-1}^2)}{(x_{n-1}^2+1)(x_{n-1}^4+2x_{n-1}^2+5)}=\frac{16(1-x_n)}{2x_n(4x_n^2+4)}=\frac{1-x_n}{x_nx_{n+1}},$$
as desired.
This shows that $d_n$ is rational; we only need to show that
$$\lim_{n\to\infty}\frac{n^2}{\ln n}\left(\frac2n-d_n\right)=2.$$
Define
$$t_n=\left(x_{n-1}-\left(1-\frac 2n\right)\right)n^2.$$
Then
\begin{align*}
t_{n+1}&=1-n^2+(n+1)^2x_n\\
&=1-n^2+(n+1)^2\left(\frac{\left(1-\frac2n+\frac{t_n}{n^2}\right)^2+1}{2}\right)\\
&=1-n^2+(n+1)^2\left(1-\frac2n+\frac2{n^2}+\frac{t_n}{n^2}-\frac{2t_n}{n^3}+\frac{t_n^2}{2n^4}\right)\\
&=\frac2n+\frac2{n^2}+t_n\left(1-\frac2{n^3}-\frac3{n^2}\right)+\frac{t_n^2}{2n^4}.
\end{align*}
So, we may show by induction on $n$ that $\{t_n-2H_{n-1}\}_n$ is bounded, where
$$H_k=\sum_{i=1}^k\frac1i$$
is the $k$th harmonic number, and so
$$x_{n-1}=1-\frac2n+\frac{2\ln n}{n^2}+O\left(\frac1{n^2}\right),$$
since $H_n\sim \ln n$. Now
\begin{align*}
\lim_{n\to\infty}\frac{n^2}{\ln n}\left(\frac2n-d_n\right)
&=\lim_{n\to\infty}\frac{n^2}{\ln n}\left(\frac2n-\frac{1-x_{n-1}}{x_{n-1}x_n}\right)\\
&=\lim_{n\to\infty}\frac{n^2}{\ln n}\left(\frac2n-\frac{\frac2n-\frac{2\ln n}{n^2}+O\left(\frac1{n^2}\right)}{x_{n-1}x_n}\right)\\
&=\lim_{n\to\infty}\frac{n^2}{x_{n-1}x_n\ln n}\left(\frac{2x_{n-1}x_n}{n}-\frac2n+\frac{2\ln n}{n^2}+O\left(\frac1{n^2}\right)\right)\\
&=\lim_{n\to\infty}\frac{2n(x_{n-1}x_n-1)}{x_{n-1}x_n\ln n}+\frac{2}{x_{n-1}x_n}+O\left(\frac{1}{x_{n-1}x_n\ln n}\right)\\
&=2-2\lim_{n\to\infty}\frac{n(1-x_{n-1}x_n)}{\ln n}.
\end{align*}
Since
$$1-x_{n-1}x_n=1-x_{n-1}+x_{n-1}(1-x_n)=O\left(\frac1n\right),$$
this limit is just $2$, as desired.
