$\ell_\infty$ is a Grothendieck space

Here's a summary of Grothendieck's argument taken from Théorème 9 on page 168 in Sur les applications linéaires faiblement compactes d'espaces du type $C(K)$, Canadian J. Math. 5 (1953), pp. 129–173. He also outlines a similar argument in Exercise 12 2) to part IV of chapter 4 of his book Topological Vector Spaces (page 229 of the 1973 Gordon and Breach edition).

Very briefly, the original proof uses a few reductions: first reducing from a general $C(K)$ space with $K$ extremally disconnected to the case of $\ell^{\infty}(S) = C(\beta S)$ (we can take $S = K_\delta$, the set underlying $K$ equipped with the discrete topology); then apply Grothendieck's criterion for weak compactness of sets of measures, which allows to reduce to the case of $\ell^{\infty}(\mathbb{N})$ and finally conclude by applying a result due to Phillips on finitely additive measures on $\mathbb{N}$ to conclude.

The result on the Grothendieck property of $\ell^\infty$ is essentially this lemma of Phillips, so it is substantially easier but at the same time it is a basic ingredient of Grothendieck's argument.

So here's the summary in more detail:

1. Recall that $C(K)$ is isometrically injective if (and only if!) $K$ is extremally disconnected. This means that $C(K)$ is $1$-complemented in every space containing it. This is proved e.g. as Theorems 4.3.6 and 4.3.7, pp. 81ff in Albiac–Kalton, Topics in Banach space theory.

   This tells us that $C(K)$ is $1$-complemented in $\ell^\infty(K)$, the space of bounded functions on $K$.

2. Let us fix a decomposition $\ell^{\infty}(K) = C(K) \oplus Z$, and assume now that we have a sequence $\mu_n$ in $M(K) = C(K)^\ast$ converging to zero in the weak$^\ast$-topology. The fixed decomposition of $\ell^{\infty}(K)$ allows us to consider this as a sequence of functionals $\mu_n \in \ell^{\infty}(K)^\ast$ satisfying $\mu_n(f) \to 0$ for all $f \in \ell^{\infty}(K)$. In other words, we can now consider $\mu_n$ as a weak$^\ast$-convergent sequence of measures in $\beta K_\delta$, the Stone–Čech compactification of the discrete space $K_\delta$.

3. We are now in position to apply Grothendieck's weak compactness criterion for $M(X)$ where $X$ is locally compact to the sequence $\mu_i$ of measures on $C(\beta K_\delta)$ (this is Theorem 2 on page 146 of G.'s article mentioned at the beginning; see also Theorem 5.3.2 on page 112 of Albiac–Kalton):

   If $\mu_i$ would not converge weakly to zero, we could find $\varepsilon \gt 0$, a subsequence $(\mu_j)_{j \in J}$ and sequence of pairwise disjoint clopen sets $U_j$ such that $|\mu_j(U_j)| \geq \varepsilon$. Since characteristic functions of clopen sets are continuous, we can translate this back to $\ell^\infty(K)$ and get a sequence of finitely additive measures $\mu_j$ on $K$ and pairwise disjoint subsets $V_j \subset K$ such that $|\mu_j(V_j)| \geq \varepsilon$. This will lead to a contradiction:

4. Define $\nu_j \in \ell^{\infty}(J)^\ast$ to be the finitely additive measure $\nu_j(B) = \mu_j\left( \bigcup_{b \in B} V_b\right)$ for all $B \subset J$. The hypothesis on the sequence $\mu_i$ implies that $\nu_j(B) \to 0$ for all $B \subset J$. With Lemma 3.3 on page 525 of Phillips, On linear transformations, Trans. Amer. Math. Soc. 48 (1940), 516–541 this leads to $$\lim_{j \to \infty} \sum_{b \in J} |\nu_j(b)| = 0,$$ so in particular we have $|\nu_j(j)| \to 0$, which contradicts $|\nu_j(j)| = |\mu_j(A_j)| \geq \varepsilon$ for all $j \in J$.

I don't know much about operator algebras, but this argument here looks like a rather direct attack on the problem and I don't think that the result would become any easier or more transparent by using operator algebra methods; hopefully someone more familiar with such reformulations will be able to contradict my assessment.