We have $$ \int_{0}^1 \int_{0}^1 \frac{p_n(x) p_n(y)}{1-xy} dx dy= \int_{0}^1 \int_{0}^1 p_n(x) p_n(y)\sum_{k=0}^\infty(xy)^k dx dy= \sum_{k=0}^\infty \left(\int_0^1 p_n(x)x^kdx\right)^2. $$ Next, integrating by parts we have $$ (-1)^n\int_0^1 p_n(x)x^kdx={k\choose n}\int_0^1x^k(1-x)^n dx=\\={k\choose n}\cdot \frac{k! n!}{(k+n+1)!}=\frac{{k\choose n}}{{k+n\choose n}}\cdot \frac1{k+n+1}. $$ We have to estimate the sum of squares of these guys over $k=0,1,\ldots$. For $k\leqslant n-1$ they are just zeros. For $k\geqslant n$ we have $$ \frac{{k\choose n}}{{k+n\choose n}}=\prod_{i=0}^{n-1}\left(1-\frac{n}{k-i+n}\right)\leqslant \prod_{i=0}^{n-1}\exp\left(-\frac{n}{k-i+n}\right)\leqslant \exp(-n^2/k). $$ The function $f(x)=\exp(-2n^2/x)(n+x+1)^{-2}$ for $x\in [n,\infty)$ has unique maximum point. Indeed, $f$ tends to 0 at infinity and $d(\log f)/dx=2n^2/x^2-2/(n+x+1)$, it equals to 0 when $n^2((n+1)/x^2+1/x)=1$, by monotonicity of $(n+1)/x^2+1/x$ it has unique positive root $x_0$ for which $n^2/x_0<1$, $x_0>n^2$. The value $f(x_0)$ is at most $n^{-4}$. The sum $\sum_{k\geqslant n} f(k)$ is therefore $O(n^{-4})+\int_n^\infty f(x)dx$. For the integral we have $$ \int_n^\infty \exp(-2n^2/x)(n+x+1)^{-2}dx\leqslant \int_n^\infty \exp(-2n^2/x)x^{-2}dx=\frac{1}{2n^2}\left(1-e^{-2n}\right). $$ So $I_n=O(n^{-2})$, and up to multiplicative factor this is sharp.

We have $$ \int_{0}^1 \int_{0}^1 \frac{p_n(x) p_n(y)}{1-xy} dx dy= \int_{0}^1 \int_{0}^1 p_n(x) p_n(y)\sum_{k=0}^\infty(xy)^k dx dy= \sum_{k=0}^\infty \left(\int_0^1 p_n(x)x^kdx\right)^2. $$ Next, integrating by parts we have $$ (-1)^n\int_0^1 p_n(x)x^kdx={k\choose n}\int_0^1x^k(1-x)^n dx=\\={k\choose n}\cdot \frac{k! n!}{(k+n+1)!}=\frac{{k\choose n}}{{k+n\choose n}}\cdot \frac1{k+n+1}. $$ We have to estimate the sum of squares of these guys over $k=0,1,\ldots$. For $k\leqslant n-1$ they are just zeros. For $k\geqslant n$ we have $$ \frac{{k\choose n}}{{k+n\choose n}}=\prod_{i=0}^{n-1}\left(1-\frac{n}{k-i+n}\right)\leqslant \prod_{i=0}^{n-1}\exp\left(-\frac{n}{k-i+n}\right)\leqslant \exp(-n^2/(n+k)). $$ The function $f(x)=\exp(-2n^2/(x+n))(n+x+1)^{-2}$ for $x\in [n,\infty)$ has the unique maximum point. Indeed, $f$ tends to 0 at infinity and $d(\log f)/dx=2n^2/(x+n)^2-2/(n+x+1)$, it equals to 0 when $n^2(1/(x+n)+1/(x+n)^2)=1$, by monotonicity it has unique positive root $x_0$ for which $n^2/(x_0+n)<1$, $x_0>n^2-n$. The value $f(x_0)$ is at most $n^{-4}$. The sum $\sum_{k\geqslant n} f(k)$ is therefore $O(n^{-4})+\int_n^\infty f(x)dx$. For the integral we have $$ \int_n^\infty \exp(-2n^2/(x+n))(n+x+1)^{-2}dx\leqslant \int_n^\infty \exp(-2n^2/(x+n))(x+n)^{-2}dx=\frac{1}{2n^2}\left(1-e^{-n}\right). $$ So $I_n=O(n^{-2})$, and up to multiplicative factor this is sharp.