By taking $t = \frac{\pi}{\sqrt{6n}}$ in formula
$$p_n = \frac{e^{nt} P(e^{-t})}{2\pi} \int_{-\pi}^{\pi} e^{-in\theta} \frac{P(e^{-t}e^{i\theta})}{P(e^{-t})} \mathrm{d}\theta,$$
conclude that
$$p_n = O\left(\frac{\exp\left(\pi\sqrt{\frac{2n}{3}}\right)}{n}\right) \quad \text{when } n \text{ tends to } +\infty.$$