First, I remark some definitions:
Given $E$ a normed vector lattice, if the norm on $E$ satisfies \begin{equation*} \|x \vee y\| = sup(\|x\|,\|y\|) \quad (x,y \in E_+), \end{equation*} then $(E,\|\cdot\|)$ is called an $M-$ normed space, and an $M-$ normed Banach lattice is called an abstract $M-$space, or $AM-$space. If the norm satisfies \begin{equation*} \|x + y \| = \|x \| + \|y\| \quad (x,y \in E_+), \end{equation*} then $(E,\|\cdot\|)$ is called an $L-$ normed space, and an $L-$ normed Banach lattice is called an abstract $L-$ space, or $AL-$space.
($E_+$ denotes the cone of positive elements on $E$).
Also, AM- and AL- spaces are dual to each other in the following sense:
The dual of each $M-$ normed space is an $AL-$ space, and the dual of each $L-$ normed space is an $AM-$ space with unit.
A famous result stated by Kakutani says the following:
Let $E$ be an $AM-$ space with unit and denote by $K$ the weak*- compact set of REAL valued lattice homomorphisms of norm one on $E$. The evaluation map $x \mapsto f$ (where $f(t) = \langle x,t \rangle \,, \, t \in K$), is an isomorphism of $E$ into $C(K)$.
Now I'm wondering, an abelian $C^*-$ algebra $\mathcal{A}$ is isomorphic to $C(K)$ where $K$ is its spectrum, a *-weak compact subset of $\mathcal{A}^*$ of norm one COMPLEX-valued homomorphisms, and with these results at hand, I would be tempted to say that any abelian $C^*-$ Algebra $\mathcal{A}$ is an $AM-$ space, as a clear consequence of Gelfand-Naimark isomorphism and Kakutani.\ This means that the dual space $\mathcal{A}$ is an $AL-$space. Indeed, for positive linear functionals $f_1,f_2 \in \mathcal{A}^*$, one obtains \begin{equation*} \|f_1 + f_2\| = (f_1 + f_2)(e)=f_1(e)+f_2(e) = \|f_1\| + \|f_2\|, \end{equation*} using that $\|f\|=f(e)$.
Can I say this? I observe that Kakutani holds for real valued maps. The point would be saying that it holds for the complexification of the lattice, but I'm not sure..
EDIT:
Ok I figured out some considerations about this question and I'm editing now to post them. In general, despite satisfying the norm- additivity on the cone of positive elements, the dual of a general(not abelian) $C^*-$ algebra is NOT an AL-space. Indeed, if I take a positive functional $f : \mathcal{A} \rightarrow \mathbb{C}$, its absolute value $|f|$ cannot be determined uniquely and this is a consequence of the not uniqueness of extension provided by the Hahn Banach theorem. So the dual space of a generic $C^*-$ algebra satisfies the AL- axiom but it is NOT an AL- space since it is just not a lattice. Indeed, if one assumed the dual of a generic $C^*-$ algebra to be AL- space, then by the duality results of AL/AM, its dual would be an AM-space, and again by Kakutani, it would be isomorphic to a C(K). But this is clearly impossible, since the bidual $\mathcal{A}^{**}$ (Enveloping von neumann algebra of $\mathcal{A}$) is clearly not commutative, in general.
Instead if we take into account a commutative $C^*-$ algebra C(X), then by kakutani and gelfand-naimark one can easily say that its self-adjoint part $\mathcal{A}_{sa}$ is isomorphic to some AM-space and in this case, its dual IS an AL-space. In fact, in this situation the uniqueness of absolute value is provided by the well-known Jordan decomposition of measure $|\mu| = \mu^+ + \mu^-$.
So the answer to the original question is : no, but its self-adjoint part is.
Hope this will be useful to someone (and correct ^_^").