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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 $\ell^1(I)$.

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 $\ell^1(I)$.

This is true, except when a cardinal invariant of of the von Neumann algebra (the maximum cardinality of a decomposition of the identity into orthogonal projections) is not a real-valued measurable cardinal.

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

It turns out that same counterexample works here. 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 $\ell^\infty(I)$, and it is easy to see that this is just pointwise convergence.

Let $\kappa$ be a real-valued measurable cardinal, and $\mu$ a $\kappa$-additive probability measure on $\mathcal{P}(\kappa)$ that vanishes on all singletons, which can be viewed as a bounded linear functional on $\ell^\infty(\kappa)$ via integration. We want to show that $\kappa$ is continuous with respect to the topology of pointwise convergence on bounded subsets of $\ell^\infty(\kappa)$. Since $|\mathbb{R}| \leq \kappa$, pointwise convergence of bounded complex-valued functions on $\kappa$ can be described by nets on directed sets of size at most $\kappa$. Let $(f_\lambda)_{\lambda \in \Lambda}$ be a bounded net such that $(f_\lambda) \to f$ and $|\Lambda| \leq \kappa$. Then by the $\kappa$-additivity of $\mu$ we have $$\int f \, d\mu = \lim_{\lambda \in \Lambda} \int f_\lambda \, d\mu.$$ Therefore, $\mu$ is continuous with respect to the topology of bounded pointwise convergence, and continuous on all strong$^*$-compact subsets of $\ell^\infty(\kappa)$. Since $\mu$ vanishes on singletons it can't have countable support, and thus can't be in $\ell^1(\kappa)$.

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

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.

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

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 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.

By a result of Akemann, the $\sigma$-strong$^*$ topology 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 precompact 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.

This is true, except when a cardinal invariant of of the von Neumann algebra (the maximum cardinality of a decomposition of the identity into orthogonal projections) is not a real-valued measurable cardinal.

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

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.

It turns out that same counterexample works here. 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 $\ell^\infty(I)$, and it is easy to see that this is just pointwise convergence.

Let $\kappa$ be a real-valued measurable cardinal, and $\mu$ a $\kappa$-additive probability measure on $\mathcal{P}(\kappa)$ that vanishes on all singletons, which can be viewed as a bounded linear functional on $\ell^\infty(\kappa)$ via integration. We want to show that $\kappa$ is continuous with respect to the topology of pointwise convergence on bounded subsets of $\ell^\infty(\kappa)$. Since $|\mathbb{R}| \leq \kappa$, pointwise convergence of bounded complex-valued functions on $\kappa$ can be described by nets on directed sets of size at most $\kappa$. Let $(f_\lambda)_{\lambda \in \Lambda}$ be a bounded net such that $(f_\lambda) \to f$ and $|\Lambda| \leq \kappa$. Then by the $\kappa$-additivity of $\mu$ we have $$\int f \, d\mu = \lim_{\lambda \in \Lambda} \int f_\lambda \, d\mu.$$ Therefore, $\mu$ is continuous with respect to the topology of bounded pointwise convergence, and continuous on all strong$^*$-compact subsets of $\ell^\infty(\kappa)$. Since $\mu$ vanishes on singletons it can't have countable support, and thus can't be in $\ell^1(\kappa)$.