Given any commutative ring $A$ let $\operatorname{Spec} A$ be the space of prime ideals of $A$. Can we interpret the Zariski topology as an initial (or final) topology with respect to some canonical maps from $\operatorname{Spec} A$ (to $\operatorname{Spec} A$)?
I am asking this because we all know that in general topology you have to know only one topology: the initial topology. (Implicit is the claim that any "canonical" topology is an initial or final topology. It would be very sad if the Zariski topology is not of this form.)
It depends on what you mean by "canonical". The Zariski topology is the weakest such that every $A\to A/I$ induces the inclusion of a closed subset. We could describe this as an initial topology by considering maps from $\operatorname{Spec} A$ to the Sierpinski space $S$ (the unique 2-point space that is $T_0$ but not Hausdorff) that send $V(I)$ to the closed point.
Generally, every topological space $X$ has the initial topology with respect to the maps $X\to S$ mapping an open set $U$ to the open point of $S$, and we can simplify this description with a nice basis for $X$. If you like thinking categorically, this can even be a good way to think of metric spaces. By Urysohn's Lemma, we can describe any normal topology as an initial topology with respect to maps to $[0,1]$. Since metric spaces are normal, every metric space is an initial topology in the category of metric spaces.
Note that the final topology is not dual in any precise sense, because the category of topological spaces is not self-dual. I doubt that it is so easy to construct a general topology in a nontrivial way as a final topology.
Addendum: You mention maps $\operatorname{Spec}A \to \operatorname{Spec}A$. If you prefer to think of only these maps, you can do the same construction as above by giving the target various appropriate topologies—this is not very useful, since describing these topologies categorically requires the Sierpinski space anyway, but I thought I'd mention it.