Let $(i, H, B)$ be an abstract Wiener space - that is, $H$ is a Hilbert space, $||\cdot ||$ is a measurable norm on $H$, $B$ is the completion of $H$ with respect to $|| \cdot||$, $i$ is the inclusion map from $H$ to $B$.
Then we can embed $B^*$ in $H$ as follows: as any measurable norm on $H$ is weaker than the norm on $H$, the restriction of any element of $B^*$ to $H$ is an element of $H^*$ - thus, by Riesz representation theorem, we can associate this element to a unique element of $H$.
So far this is what I have (don't know if this will help):
A characterization of $B^*$ is: $$ B^* = \left\{x \in H: \sup_{y \in H, y \neq 0} \frac{\langle x, y\rangle}{||y||} < \infty \right\}$$
(Note that $\langle\cdot, \cdot \rangle$ is the inner product on $H$ and $||\cdot||$ is the measurable norm on $H$. Also, if this quantity exists and is finite then it is the $B^*$ norm of $x$)
Apparently, $B^*$ is dense in $H$, but it is not apparent to me why this is true.
Write $i^*$ for the natural map from $B^*$ into $H$. It is indeed the adjoint of $i$ since for $f \in B^*$, $h \in H$ we have $\langle i^*f, h \rangle_H = f(ih)$.
Suppose $h_0 \in (i^*(B^*))^\perp$, so $\langle i^* f, h_0 \rangle =0$ for all $f \in B^*$. This means $f(i h_0) = 0$ for all $f \in B^*$, so by the Hahn-Banach theorem, $i h_0 = 0$ which means $h_0 = 0$. This implies $i^*(B^*)$ is dense in $H$.