It's not sequential because its closed subspace $M[0,1] = (C[0,1])^\ast$ is not sequential.
Here is an example of a set $A \subset M[0,1]$ that is sequentially $\tau_v$-closed but not $\tau_v$-closed:
Consider sequence of functions $f_n \in C[0,1]$, $\Vert f \Vert_{C[0,1]} = 1$ and $\operatorname{span} \{f_n\}$ is norm-dense in $C[0,1]$, and take $$A := \bigcup_{n \ge 1} \left\{ \mu : \intop f_n d \mu = n \right\},$$
On the one hand, it's sequentially $\tau_v$-closed. Indeed, any $\tau_v$-convergent sequence is bounded in the total variation norm by the uniform boundedness principle, and $A$ intersects any ball at only finitely many closed hyperplanes. On the other hand, $0 \notin A$ but $0$ is contained in the $\tau_v$-closure of $A$ because any basic $\tau_v$-neighborhood of $0$ of the form $\left\{ \left| \intop g_i d \mu \right| < \varepsilon, i = 1, \dots, n \right\}$ intersects $\left\{ \intop f_m d \mu = m \right\}$ for any $f_m \notin \operatorname{span} \{ g_i\}$.
It's Lusin.
One can "encode" a measure on $\mathbb{R}$ by a sequence of compatible finite measures on $[-n, n]$. Now, on finite measures on $[-n, n]$ the weak topology can be strenghthened to a Polish space topology by adding total variation norm balls around $0$ as additional open sets. Here we are using the fact that on balls the weak topology becomes metrizable. For example, one can use the following metric: $$\rho(\mu, \nu) := \sum_m \frac{1}{2^m} \arctan \left| \intop f_m d (\mu - \nu) \right| + \left| \Vert \mu \Vert - \Vert \nu \Vert \right|$$ where $\Vert \cdot \Vert$ is the total variation norm and $f_m$ is a dense sequence of functions of norm $1$.
The condition that the measures on $[-n, n]$ are compatible introduces a closed subset in the countable product of spaces of measures on $[-n, n]$. So this way we can cook up a Polish space topology on $M(\mathbb{R})$.