The potential energy is $U\left(x\right) = kx^4/4$ since $-d/dx\left(kx^4/4\right) = -kx^3 = F$, and the energy
$$
E = \frac{1}{2}m\left(\frac{dx}{dt}\right)^2 + \frac{1}{4}kx^4
$$
is conserved.
From the above you can show that
$$
\begin{eqnarray}
dt &=& \pm \ dx \sqrt{\frac{m}{2E}}\left(1-\frac{k}{4E}x^4\right)^{-1/2} \\
&=& \pm \ dx \sqrt{\frac{2m}{k}} \ A^{-2} \left[1-\left(\frac{x}{A}\right)^4\right]^{-1/2}
\end{eqnarray}
$$
where the amplitude $A = \left(4E / k\right)^{1/4}$ can be found from setting $dx/dt = 0$ in the expression for the energy and solving for $x$.
The period is then
$$
\begin{eqnarray}
T &=& 4 \sqrt{\frac{2m}{k}} \ A^{-2} \int_0^A dx \left[1-\left(\frac{x}{A}\right)^4\right]^{-1/2}
\\
&=& 4 \sqrt{\frac{2m}{k}} \ A^{-1} \int_0^1 du \left(1-u^4\right)^{-1/2} \\
&=& \left(4 \sqrt{\frac{2m}{k}} I\right) A^{-1} \\
&\propto& A^{-1}
\end{eqnarray}
$$
where $u = x/A$ and $I = \int_0^1 du \left(1-u^4\right)^{-1/2} \approx 1.31$ (see this).
You can repeat the above for a more general potential energy $U\left(x\right) = \alpha \left|x\right|^n$, where you should find that
$$
dt = \pm \ dx \sqrt{\frac{m}{2\alpha}} \ A^{-n/2} \left[1-\left(\frac{\left|x\right|}{A}\right)^n\right]^{-1/2}
$$
and
$$
\begin{eqnarray}
T_n &=& \left(4 \sqrt{\frac{m}{2\alpha}} I_n\right) A^{1-n/2} \\
&\propto& A^{1-n/2}
\end{eqnarray}
$$
where
$$
I_n = \int_0^1 du \left(1-u^n\right)^{-1/2}
$$
can be evaluated in terms of gamma functions (see this).
This is in agreement with the above for $\alpha = k/4$ and $n=4$, and with Landau and Lifshitz's Mechanics problem 2a of section 12 (page 27), where they find that $T_n \propto E^{1/n-1/2} \propto A^{1-n/2}$.
