[As requested by the asker, I edited this answer to include more background information on the topologies I mention.]

The topology you refer to in your question is the strong$^*$ (operator) topology, defined by the seminorms
$$a \mapsto \|a \xi\| \quad \text{ and } \quad a \mapsto \|a^* \xi\|$$
for $\xi \in \mathcal{H}$. The $\sigma$-strong$^*$ (operator) topology is defined by the seminorms
$$a \mapsto \sum_{n = 1}^\infty \|a \xi_n\| \quad \text{ and } \quad a \mapsto \sum_{n = 1}^\infty \|a^* \xi_n\|$$
for square-summable sequences $(\xi_n)_{n=1}^\infty$. Similarly, the strong operator topology is defined by the seminorms
$$a \mapsto \|a \xi\|$$
and the $\sigma$-strong (operator) topology is defined by the seminorms
$$a \mapsto \sum_{n = 1}^\infty \|a \xi_n\|.$$
The weak operator topology is defined by the seminorms
$$a \mapsto |\langle a \xi \,\vert\, \eta \rangle|$$
and the $\sigma$-weak (operator) topology, which coincides with the weak$^*$ topology, is defined by the seminorms
$$a \mapsto \sum_{n = 1}^\infty |\langle a \xi_n \,\vert\, \eta_n \rangle|.$$
It's a standard result that all of the non-$\sigma$ topologies have the same continuous linear functionals and all of the $\sigma$ topologies have the same continuous linear functionals. Since the $\sigma$ and non-$\sigma$ variants of each topology agree on bounded sets, it follows from the Krein-Smulian Theorem applied to the $\sigma$-weak topology that they all have the same closed convex sets. 

The advantage of the $\sigma$ versions of the topologies is that every $*$-isomorphism of von Neumann algebras is a homeomorphism for any of those topologies, which is not true for the non-$\sigma$ versions. Their defining seminorms can also be defined using the predual, with no reference to a Hilbert space, e.g. the $\sigma$-strong$^*$ topology is given by the seminorms
$$a \mapsto \varphi(a^* a)^{1/2} \quad \text{ and } \quad a \mapsto \varphi(a a^*)^{1/2}$$
for $\varphi \in \mathcal{M}_*^+$. Hence your question is a well-defined question about the $\sigma$-strong$^*$ topology of an abstract von Neumann algebra.

If $\mathcal{M}$ is a von Neumann algebra, then any linear functional continuous on compact subsets of the $\sigma$-strong$^*$ topology is sequentially $\sigma$-strong$^*$ continuous. Let $\varphi$ be a linear functional on $\mathcal{M}$ that is $\sigma$-strong$^*$-continuous on compact sets. If $(x_n) \to 0$ is $\sigma$-strong$^*$-convergent, then $K = \overline{\{ x_n : n \in \mathbb{N} \}}$ is $\sigma$-strong$^*$-compact. Continuity of $\varphi$ on $K$ implies that $(\varphi(x_n)) \to \varphi(0)$. Therefore $\varphi$ is sequentially continuous.

If $\mathcal{M}_*$ is separable, then the $\sigma$-strong$^*$ topology is metrizable on bounded parts of $\mathcal{M}$. Choose a dense subset $(\psi_n)_{n=1}^\infty$ of the unit ball of $\mathcal{M}_*$. Then
$$d(x, y) = \sum_{n = 1}^\infty \psi_n((x - y)^* (x - y))^{1/2} + \sum_{n = 1}^\infty \psi_n((x - y) (x - y)^*)^{1/2}$$
defines a metric for the strong$^*$ topology on the unit ball of $\mathcal{M}$. More generally, if $\mathcal{M}$ is countably decomposable and $\psi$ is a faithful normal state on $\mathcal{M}$, then
$$d(x, y) = \psi((x - y)^* (x - y))^{1/2} + \psi((x - y) (x - y)^*)^{1/2}$$
defines a metric for the strong$^*$ topology on the unit ball of $\mathcal{M}$.

Since sequential continuity is equivalent to continuity on metric spaces, this gives an affirmative answer to your question when $\mathcal{M}$ is countably decomposable. Similar results hold for the other operator topologies.

More generally, it is pretty likely that sequential $\sigma$-strong$^*$ continuity doesn't imply $\sigma$-strong$^*$ continuity. I didn't verify that the details carry over to the $\sigma$-strong$^*$ topology, but [Matthias Neufang has shown](http://people.math.carleton.ca/~mneufang/JOT_60_08.pdf) that sequential $\sigma$-weak continuity is equivalent to $\sigma$-weak continuity if and only if the decomposability number of $\mathcal{M}$ is not a measurable cardinal, with $\ell^\infty(\kappa)$ for a measurable cardinal $\kappa$ being a counterexample.

The $\sigma$-strong$^*$ topology is quasicomplete (complete on bounded sets), so the totally bounded sets are precisely the relatively compact sets, and it agrees with the Mackey topology on bounded sets. The Grothendieck compactness criterion for the Mackey topology on the dual of a Banach space is that a set $E \subseteq \mathcal{M}$ is relatively compact if and only if
$$\lim_{n \to \infty} \sup_{x \in E} |\varphi_n(x)| = 0$$
for all weakly null sequences $(\varphi_n)$ in $\mathcal{M}_*$. It might be possible to use the decomposition of $\mathcal{M}^*$ into normal and singular parts and the weak sequential completeness of $\mathcal{M}_*$ to answer your question for all von Neumann algebras.