Nonlinear spring $F=-kx^3$

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 $T_n \propto E^{1/n-1/2} \propto A^{1-n/2}$.

You can use Dimensional analysis to get the relationship between Time period(T) and Amplitude.(A)

$$F=-kx^3$$

$$MLT^{-2} = K L^3$$

This would imply that $T^{-2}\propto L^2$ i.e. $TL=$ Constant

$T$ is inversely proportional to $L$

$L$ can be taken as amplitude also.