= Solution
Newton's equation and its energy <integral> are
$$
m\ddot x=-V'(x),\qquad \frac12m\dot x^2+V(x)=E.
$$
Thus
$$
T=\sqrt{2m}\int_{x_-}^{x_+}\frac{dx}{\sqrt{E-V(x)}}.
$$
Let $a=-V''(x_+^*)>0$. Near the maximum,
$$
V(x)=E^*-\frac a2(x-x_+^*)^2+O(|x-x_+^*|^3),
$$
and the nearby turning point has distance $\epsilon\sim\sqrt{2E^*\delta/a}$ from $x_+^*$. The singular part of the period is an $\operatorname{arcosh}$ <integral>:
$$
\sqrt{2m}\sqrt{\frac2a}\int_\epsilon^c\frac{dy}{\sqrt{y^2-\epsilon^2}}
=\sqrt{\frac ma}\log(1/\delta)+O(1).
$$
Hence
$$
T=\sqrt{\frac m{-V''(x_+^*)}}\{\log(1/\delta)+O(1)\}.
$$
Physically, the limiting orbit approaches the <unstable equilibrium> at the barrier top with vanishing speed and spends arbitrarily long there.
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