The finitely support probability measures on $\mathbb Z$ are all members of $\ell^1(\mathbb Z)$. So we could ask a slightly more general question:

What is the sequential closure of (the probability measures in) $\ell^1(\mathbb Z)$ in $\ell^\infty(\mathbb Z)^*$?

If $(a_n)$ is a sequence in $\ell^1(\mathbb Z)$ which converges weak$^*$ in $\ell^\infty(\mathbb Z)^*$, then $f(a_n)$ is a Cauchy sequence for all $f\in \ell^\infty(\mathbb Z)$. Now, $\ell^1(\mathbb Z)$ is weakly sequentially complete (*) and so there is $a\in\ell^1(\mathbb Z)$ with $f(a_n) \rightarrow f(a)$ for each $f\in\ell^\infty(\mathbb Z)$. In other words, the weak$^*$ limit of $(a_n)$ actually lives in $\ell^1(\mathbb Z)$. So the answer is "no" in a strong sense.

(*) All preduals of a von Neumann algebra are weakly sequentially complete (see Takesaki I, Chapter III, Corollary 5.2). There is presumably an easy proof for $\ell^1$, but I don't know a reference off the top of my head.

**Edit:** The following is prompted by Bill Johnson's comment (and reveals nothing so much as my own ignorance!) A Banach space $E$ is said to have the Schur Property if weakly convergent sequences are norm convergent-- Schur originally showed that $\ell^1$ has the Schur Property. We say that $E$ is weakly sequentially complete (WSC) if whenever $(x_n)$ is weakly Cauchy (meaning that for each $f\in E^*$, the scalar sequence $(f(x_n))$ is Cauchy) then $(x_n)$ is weakly convergent. The Schur property implies WSC-- if $(x_n)$ is weakly Cauchy, then for any increasing sequences $(n_k)$ and $(m_k)$, the sequence $(x_{n_k} - x_{m_k})_k$ is weakly-null, hence norm-null, and hence $(x_n)$ is norm Cauchy. You can find all this in Section 2.3 of Albiac and Kalton's book. For a general measure $\mu$, the space $L^1(\mu)$ is WSC, but doesn't necessarily have the Schur property-- see Section 5.2 of this book.