Why not $\sf ZFC+[V=HOD]$ as the standard set theory?

It implies the existence of a definable global choice and well-order, and it is compatible with all large cardinal axioms extending $\sf ZFC$, so per maximality standpoint we are not losing anything, and more we are gaining a stronger notion of choice and more theorization and expressibility. Also, $\sf V=HOD$ is equivalent to saying that the parameter free definable sets forms an elementary substructure of the universe of sets, so it is saying that whatever is done by sets can be done by using parameter free definable sets. It's only among the latter ones that concrete sets can thrive, and accordingly are of the most convincing kind of sets intuitively speaking, so seeing that all work done in set theory can be done using them is a very strong feature to hold to, intuitively speaking. If the exposition of $\sf V=HOD$ is seen as being complex, it can be expressed metatheoretically in a simple manner by adding to $\sf ZFC$ an $\omega$-rule of definability that is:

$\textbf{Definability: }$ if $\phi_1,\phi_2, \phi_3,...$ are all $\sf FOL(=,\in)$ formulas in which only symbol "$y$" occurs free, and "$y$" never occur bound, and that doesn't use the symbol "$x$", and $\psi$ is a formula in which only symbol "$x$" occurs free, and "$x$" never occur bound; then:

$\underline {[i=1,2,3,...; \\ \forall x \, (x=\{y \mid \phi_i\} \to \psi)]} \\ \forall x: \psi$

In English: if a parameter free formula holds for all parameter free definable sets, then it holds for all sets.

This was proved by Hamkins to be equivalent over $\sf ZFC$ to the set theoretic axiom $\sf V=HOD$

So, it meets all intuitive and technical qualifications for it to be there!?

clearly true. Not a formal gimmick. $\endgroup$5more comments