For a rotating ball you should use for energy $$E=\frac{1}{2}mv^2+\frac{1}{2}I\omega^2+mgh$$ where $I$ is the moment of inertia which for a sphere is $$I=\frac{2}{5}mr^2$$ where $r$ is the radius of the sphere, and $\omega$ it's his angular velocity which is related to the velocity of the center of mass via the equation $$\vec{r}\times \vec{\omega}=\vec{v}$$ in you case if the sphere moves along a single direction you may write $\omega=vr$ where $r$ is the radius, maybe you also want to check if friction is a relevant factor. If not you will get $$g=\frac{7}{10}\frac{v^2_{final}}{l\sin(\alpha)}$$ where l is the distance the sphere travelled on the plane and $\alpha$ is the angle of inclination of the plane
$UPDATE$ (due to comment)
In the presence of friction you will have a relation of the type: $$E(t)-E(0)=\epsilon(t)\rightarrow \frac{7}{10}mv^2(t)-mgl\sin(\alpha)=\epsilon(t)$$ energy is not conserved and the difference between initial and final energy is an unkown function of time,thus you will have $$v^2(t)=\frac{10}{7}g\sin(\alpha)l(t)+\frac{10\epsilon(t)}{7m}$$ so if you plot $v^2$ for different values of $l(t)$ (basically at different times) you should get a line with coefficient $\frac{10}{7}g\sin(\alpha)$ if the contribution of friction, which i stress is an unkown funtion of time, it's strong than it will be probably something else than a line since $\epsilon=\epsilon(t,l(t),v(t),..)$ you dont really know, if you get a line it might not be crossing the origin since for weak it might be you still have a contribution from friction, but in the case of a linear relation you can calculate the angular coefficent $$\frac{10}{7}g\sin(\alpha)$$ to estimate $g$
