Let $v$ be a vector field. We say that $v$ is covariantly constant iff
\begin{equation} \nabla v=0, \end{equation} where $\nabla v= (\partial_j v^i+\Gamma^i_{jk} v^k) \ \partial_i \otimes dx^j$, i.e. the total covariant derivative of $v$ is zero. Now let $v=\omega^{\sharp}$ be a rised one-form. Therefore \begin{equation} \nabla v=\nabla(\omega^{\sharp})=\ldots \end{equation} is there a way to simplify this expression? I would like that expresion to be in terms of $\omega$. I've tried simply $\nabla \omega = 0$, but it doesn't work.
I'm not sure what you mean by "it doesn't work". It works, since you have the metric (hence musical isomorphism) is parallel.
Start with the Leibniz rule: $$ \begin{align*} X(\omega(Y))&=(\nabla_X\omega)(Y)+\omega(\nabla_XY)\\ X\langle v,Y\rangle&=\langle\nabla_Xv,Y\rangle+\langle v,\nabla_XY\rangle \end{align*} $$ Equating them gives $$ (\nabla_X\omega)(Y)=\langle\nabla_Xv,Y\rangle $$ so $\nabla_Xv=(\nabla_X\omega)^\sharp$, i.e., $\nabla(\omega^\sharp)=(\nabla\omega)^\sharp$.