According to Mathematica your polynomials satisfy the recurrence relation
$$
(2 n+1) p(n) \left(n^2+3 n-2 x+2\right)+p(n+1) \left(-4 n^3-18 n^2+4 n x-27 n+2 x-14\right)+(2 n+3) (n+2)^2 p(n+2)=0
$$
with initial conditions $p_1=1+2x,p_2=1-x(x-13)/6$. The solution to this equation is
$$
p_n(x)=\frac{512 n (x-1) x^2 \, _3F_2\left(-\frac{1}{2},\frac{1}{2}-\frac{1}{2} \sqrt{8 x+1},\frac{1}{2} \sqrt{8 x+1}+\frac{1}{2};\frac{1}{2},1;1\right)}{(2 n+1) \left(\sqrt{8 x+1}-3\right) \left(\sqrt{8 x+1}-1\right)^2 \left(\sqrt{8 x+1}+1\right)^2 \left(\sqrt{8 x+1}+3\right)}+\frac{256 (x-1) x^2 \, _3F_2\left(-\frac{1}{2},\frac{1}{2}-\frac{1}{2} \sqrt{8 x+1},\frac{1}{2} \sqrt{8 x+1}+\frac{1}{2};\frac{1}{2},1;1\right)}{(2 n+1) \left(\sqrt{8 x+1}-3\right) \left(\sqrt{8 x+1}-1\right)^2 \left(\sqrt{8 x+1}+1\right)^2 \left(\sqrt{8 x+1}+3\right)}+\frac{256 (x-1) x^2 \cos \left(\frac{1}{2} \pi  \sqrt{8 x+1}\right) \Gamma \left(n+\frac{1}{2} \sqrt{8 x+1}+\frac{3}{2}\right) \Gamma \left(n-\frac{1}{2} \sqrt{8 x+1}+\frac{3}{2}\right) \, _4F_3\left(1,n+\frac{1}{2},n-\frac{1}{2} \sqrt{8 x+1}+\frac{3}{2},n+\frac{1}{2} \sqrt{8 x+1}+\frac{3}{2};n+\frac{3}{2},n+2,n+2;1\right)}{\pi  (2 n+1) \left(\sqrt{8 x+1}-3\right) \left(\sqrt{8 x+1}-1\right)^2 \left(\sqrt{8 x+1}+1\right)^2 \left(\sqrt{8 x+1}+3\right) \Gamma (n+2)^2}+\frac{1024 n (x-1) x^3}{(2 n+1) \left(\sqrt{8 x+1}-3\right) \left(\sqrt{8 x+1}-1\right)^2 \left(\sqrt{8 x+1}+1\right)^2 \left(\sqrt{8 x+1}+3\right)}+\frac{512 (x-1) x^3}{(2 n+1) \left(\sqrt{8 x+1}-3\right) \left(\sqrt{8 x+1}-1\right)^2 \left(\sqrt{8 x+1}+1\right)^2 \left(\sqrt{8 x+1}+3\right)}+\frac{512 n (x-1) x^2}{(2 n+1) \left(\sqrt{8 x+1}-3\right) \left(\sqrt{8 x+1}-1\right)^2 \left(\sqrt{8 x+1}+1\right)^2 \left(\sqrt{8 x+1}+3\right)}+\frac{256 (x-1) x^2}{(2 n+1) \left(\sqrt{8 x+1}-3\right) \left(\sqrt{8 x+1}-1\right)^2 \left(\sqrt{8 x+1}+1\right)^2 \left(\sqrt{8 x+1}+3\right)}
$$

For $n\to\infty$, Mathematica claims that this becomes
$$
p_\infty(x)=\frac{1}{2} \left(\, _3F_2\left(-\frac{1}{2},\frac{1}{2}-\frac{1}{2} \sqrt{8 x+1},\frac{1}{2} \sqrt{8 x+1}+\frac{1}{2};\frac{1}{2},1;1\right)+2 x+1\right)
$$
which solves $p(x)=x$ at around
$$
x=-0.573825523080029241015952733\dots
$$