Symplectic orthogonality and projective duality: how do they work together?
Corrected (Partial) Answer
Given a smooth variety $X\subset \mathbb{P}V$ that is the projectivization of a smooth punctured cone $\hat X\subset V\setminus\{0\}$, the space $X_\bullet$ that the OP defines is a variety in $\mathbb{P}V$ described as $$ X_\bullet = \bigl\{[v]\in\mathbb{P}V\ \bigl|\ \exists w\in \hat X,\ \ \omega(v,T_w\hat X) = 0\ \bigr\} \subset \mathbb{P}V. $$ The OP's question then is: When is $X_\bullet = X$? (In this case, say $X$ is $\omega$-self-dual.) [Note that the question of when $X_\bullet$ is smooth or has the same dimension as $X$ is not really a symplectic question, since the classical dual $X^*\subset\mathbb{P}V^*$ is equivalent to $X_\bullet$ under the correlation defined by $\omega$, and $X^*$ does not depend on the symplectic structure.]
I don't have a complete answer, but here are a couple of observations:
If $X$ is $\omega$-self-dual, we must have $\dim X \ge n{-}1$, and, if $\dim X = n-1$, then $X = \mathbb{P}L$, where $L\subset V$ is a Lagrangian subspace of $V$. The reason for this is that, if $\dim X = k$, then the dimension of $T_w\hat X$ is $k+1$, so its symplectic orthogonal $(T_w\hat X)^\perp$ has dimension $2n-k-1$ and so $\mathbb{P}\bigl((T_w\hat X)^\perp\bigr)$ has dimension $2n-k-2$ and is contained in $X_\bullet$. Thus, if $X_\bullet = X$ (or, more generally, if they just have the same dimension), then $k\ge 2n-k-2$, so $k\ge n{-}1$. If equality holds then $X = X_\bullet = \mathbb{P}\bigl((T_w\hat X)^\perp\bigr)$, so $X = \mathbb{P}L$ for the Lagrangian space $L = (T_w\hat X)^\perp = T_w\hat X$.
If $\dim X = 2n{-}2$ (i.e., $X$ is a hypersurface), then $X$ must be a quadric hypersurface (since those are the only smooth hypersurfaces whose duals are smooth), but not every quadric hypersurface is $\omega$-self-dual when $n>1$: The set of smooth quadric hypersurfaces in $\mathbb{P}V$ has dimension $n(2n{+}1)-1$ while the space of $\omega$-self-dual quadric hypersurfaces only has dimension $n(n{+}1)$. In fact, it is not hard to show that if $V = L_+\oplus L_-$ is a splitting of $V$ into two transverse Lagrangian subspaces $L_+$ and $L_-$, then the quadric hypersurface defined by $$ X = \bigl\{ [v_+ + v_-]\in\mathbb{P}V\ \bigl|\ \omega(v_+,v_-) = 0\ \bigr\} $$ is $\omega$-self-dual, and every $\omega$-self-dual quadric hypersurface is of this form for some splitting $V = L_+\oplus L_-$.
Varieties having the same dimension as their duals have been classified by Ein (see http://link.springer.com/article/10.1007%2FBF01391495). They are:
_hypersurfaces,
_$\mathbb{P}^1 \times \mathbb{P}^{n-1} \subset \mathbb{P}^{2n-1}$;
_$G(2,5) \subset \mathbb{P}^9$,
_$\mathbb{S}_{10} \subset \mathbb{P}^{15}$.
EDIT : this results holds if one assumes the variety non-degenerate (that is non-included in a hyperplane).