[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 $\varphi \in \mathcal{M}_*^+$.

Both the strong$^*$ and $\sigma$-strong$^*$ topologies are quasicomplete, meaning that all closed and bounded sets are complete, and hence total boundedness is equivalent to precompactness. Also, the notion of boundedness (in the topological vector space sense) agrees with the ordinary notion of norm boundedness for all of these topologies. For the $\sigma$ topologies for whom the continuous linear functionals are $\mathcal{M}_*$, this follows from the Banach-Mackey Theorem. For the non-$\sigma$ variants it also follows because their continuous linear functionals are dense in $\mathcal{M}_*$.

Since a net $(a_\lambda)$ is Cauchy (resp. converges to $a$) in the strong$^*$-topology if and only if $(a_\lambda)$ and $(a_\lambda^*)$ are Cauchy (resp. converge to $a$ and $a^*$) in the strong operator topology, we just need to show quasicompleteness for the strong operator topology. But a bounded Cauchy net in the strong operator topology has a limit by the Uniform Boundedness Principle. This also shows that the $\sigma$-strong and $\sigma$-strong$^*$ topologies are quasicomplete, because they are finer than their corresponding non-$\sigma$ versions but agree on bounded sets. 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.

Matthias Neufang has shown that sequential $\sigma$-weak continuity is equivalent to $\sigma$-weak continuity if and only if the decomposability number of $\mathcal{M}$ is not a real-valued measurable cardinal, with $\ell^\infty(\kappa)$ for a real-valued measurable cardinal $\kappa$ being a counterexample. The arguments should generalize to the $\sigma$-strong$^*$ topology, because they reduce to the commutative case, and then reduce to a direct sum of finite measure spaces, which give countably decomposable von Neumann algebras.

In the case of $\ell^\infty(I)$, your question has an affirmative answer. First, some preliminaries on various topologies on $\ell^\infty(I)$. This has some overlap with the last question I answered. The Mackey topology is the finest vector topology on $\ell^\infty(I)$ such that the continuous linear functionals are $\ell^1(I)$, and it is given by convergence on the absolutely convex weakly compact subsets of $\ell^1(I)$. The bounded weak$^*$ topology (or the equicontinuous weak$^*$ topology in the duals of non-normed spaces) is the finest vector topology that agrees with the weak$^*$ topology on bounded sets, and it is given by convergence on the absolutely convex norm compact sets of $\ell^1(I)$. Its continuous linear functionals are also $\ell^1(I)$. Since $\ell^1(I)$ has the Schur property, these topologies coincide on $\ell^1(I)$.

By a theorem of Akemann, the strong$^*$ and Mackey topologies agree on bounded sets, so the strong$^*$ topology agrees with the weak$^*$ topology on bounded subsets of $\ell^\infty(I)$. The Grothendieck Completeness Theorem applied to $\ell^1(I)$ says that the completeness of $\ell^1(I)$ implies that every linear functional on $\ell^\infty(I)$ that is weak$^*$-continuous on the absolutely convex weak$^*$-compact subsets of $\ell^\infty(I)$ (or just the unit ball) is in $\ell^1(I)$.

Since every commutative von Neumann algebra is a product of copies of $\mathbb{C}$ and $\mathrm{L}^\infty([0, 1])$, whose strong$^*$ topologies are all sequential, one might expect this to generalize by analyzing weak compactness in the $\ell^1$ sum of the preduals with respect to weak compactness in the individual components, but I can't see how to make it follow easily from any other theorem.