Does taking $\nabla\times$ infinity times from an arbitrary vector exists? Is it possible to get the value of:
\begin{equation}
\underbrace{\left[\nabla\times\left[\nabla\times\left[\ldots\nabla\times\right.\right.\right.}_{\infty\text{-times taking curl operator}}\mathbf{V}\left.\left.\left.\right]\right]\ldots\right] = ?
\end{equation}
For any possible values of vector $\mathbf{V}$.

For a general vector field $\mathbf{V}$, this sequence need not converge.  Consider for example $\mathbf{V} = (e^{x-y}, e^{x-y}, 0)$.  We have $\nabla \times \mathbf{V} = \mathbf{W} = (0,0,2 e^{x-y})$, and $\nabla \times \mathbf{W} = -2 \mathbf{V}$, and the cycle then repeats with a factor of $-2$.
It's worth noting that this operation is "unnatural" from the point of view of differential forms; curl is really the $d$ operator taking 1-forms to 2-forms; it's just that in $\mathbb{R}^3$, both 1-forms and 2-forms can be viewed as vector fields.  But the result of the curl operator is a 2-form, which is not something that it makes sense to take the curl of.  The other relations you cite in your comment don't suffer from this problem: grad takes 0-forms to 1-forms (so curl grad makes sense) and div takes 2-forms to 3-forms (so div curl makes sense), and they are both special cases of the fact that $d^2 = 0$.
