This answer combines @AntonioVargas and @GregMartin results.

Let us start from the approximation in Laplace's method (https://en.wikipedia.org/wiki/Laplace%27s_method)

$$\int_a^b h(x)e^{Mg(x)}dx \approx \sqrt{\frac{2\pi}{M\|g''(x_0)\|}}h(x_0)e^{Mg(x_0)}$$

and rewrite the general integral in the question as

$$\int_0^1 \frac{4x^2}{1+x^2}e^{n\log\left(\dfrac{x(1-x)^2}{2}\right)} dx$$

so we may identify

$$h(x)=\dfrac{4x^2}{1+x^2}$$

$$g(x)=\log\left(\dfrac{x(1-x)^2}{2}\right)$$

$$g'(x)=\dfrac{1-3x}{x(1-x)}$$

$$g''(x)=-\dfrac{1-2x+3x^2}{x^2(1-x)^2}$$

The position of the unique global maximum of $g(x)$ is obtained from $g'(x_0)=0$

$$x_0=\frac{1}{3}$$

Substituting into $h(x)$, $g(x)$ and $g''(x)$, we obtain

$$h(x_0)=h\left(\dfrac{1}{3}\right)=\frac{4\left(\dfrac{1}{3}\right)^2}{1+\left(\dfrac{1}{3}\right)^2}=\frac{2}{5}$$

$$g(x_0)=log\left(\dfrac{x_0(1-x_0)^2}{2}\right) = log\left(\frac{2}{27}\right)$$

$$g''(x_0)=g''\left(\dfrac{1}{3}\right)=-\frac{27}{2}$$

Finally,

$$\sqrt{\frac{2\pi}{M\|g''(x_0)\|}}h(x_0)e^{Mg(x_0)} = \sqrt{\dfrac{2\pi}{n\frac{27}{2}}}\frac{2}{5}e^{n\log\left(\dfrac{2}{27}\right)} = \frac{4}{15}\sqrt{\dfrac{\pi}{3n}}\left(\dfrac{2}{27}\right)^n,$$

so

$$\int_0^1 \frac{x^{n+2}(1-x)^{2n}}{2^{n-2}(1+x^2)}dx \sim \frac{4}{15}\sqrt{\dfrac{\pi}{3n}}\left(\dfrac{2}{27}\right)^n$$

for large $n$.

The asymptotic quality is therefore

$$\theta = \lim_{n \to \infty} -\frac{\log\left(\dfrac{p_n}{q_n}-\pi\right)}{\log(q_n)} = \lim_{n \to \infty} -\frac{\log\left( \dfrac{4}{15}\sqrt{\dfrac{\pi}{3n}}\left(\dfrac{2}{27}\right)^n \right)}{\log( (2e^3)^n)} = \frac{3log(3)-log(2)}{3+log(2)} \approx 0.7 < 1,$$

which is low.