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 $\mathbf{V}$.

Two applications of $\nabla$ yield $\nabla \times (\nabla \times F) = -\nabla^2 F + \nabla(\nabla \cdot F)$. Why? Well, setting $F = \sum_i F_i e_i$ where $e_i$ is the standard cartesian frame of $\mathbb{R}^3$ allows the formula:
$$ (\nabla \times F)_k = \sum_{ij} \epsilon_{ijk} \partial_i F_j $$
Curling once more,
$$ [\nabla \times (\nabla \times F)]_m = \sum_{kl}\epsilon_{klm}\partial_k\sum_{ij} \epsilon_{ijl} \partial_i F_j $$
But, the antismmetric symbol is constant and we can write this as
$$ [\nabla \times (\nabla \times F)]_m = \sum_{ijkl}\epsilon_{klm}\epsilon_{ijl} \partial_k \partial_i F_j $$
A beautiful identity states:
$$ \sum_{l}\epsilon_{klm}\epsilon_{ijl} = -\sum_{l}\epsilon_{kml}\epsilon_{ijl} =
-\delta_{ki}\delta_{mj}+\delta_{kj}\delta_{mi}$$
Hence,
$$ [\nabla \times (\nabla \times F)]_m = \sum_{ijk}(-\delta_{ki}\delta_{mj}+\delta_{kj}\delta_{mi}) \partial_k \partial_i F_j = \sum_i [-\partial_i^2F_m+\partial_m(\partial_iF_i)]$$
and the claim follows since $m$ is arbitrary. Now, let's try for 3:
$$ \nabla \times (\nabla \times (\nabla \times F)) = \nabla \times \bigl[-\nabla^2 F + \nabla(\nabla \cdot F)\bigr] =\nabla \times (-\nabla^2 F)$$
I used the curl of a gradient is zero. This need not be trivial, take $F = <x^2z,0,0>$ as an example. I suppose I could have shot for a four-folded curl by doubly applying the identity.
$$ \nabla \times (\nabla \times (\nabla \times (\nabla \times F))) =?$$
Set $G = -\nabla^2 F $ since we know the gradient term vanishes,
$$ \nabla \times (\nabla \times G) = -\nabla^2 G + \nabla \cdot G = \nabla^2(\nabla^2 F)-\nabla [\nabla \cdot (\nabla^2 F)]$$
So, there's the four-folded curl. Well, I see no reason this terminates. I guess you can give it a name. I propose we call (ordered as the edit indicates) $ \nabla \times \nabla \times \cdots \times \nabla = \top$
