I am looking for a reference to answer the question in the title. Let me try to clarify a little what I mean:
If a single sheaf $\mathscr F$ has a resolution $\mathscr G^\bullet$ by not necessarily injective objects, then the usual cohomology of $\mathscr F$ is isomorphic to the hypercohomology of $\mathscr G^\bullet$: $$ H^i(X, \mathscr F) \cong \mathbb H^i(X,\mathscr G^\bullet). $$
Now, if one was starting with a complex of sheaves $\mathscr F^\bullet$ and a "resolution" thereof, i.e. a complex of complexes $(\mathscr G^\bullet)^\bullet$, then one should touch on a concept that could be called hyper-hypercohomology.
Yet, I never heard of its existence and I'm pretty sure it does not give you anything new, as soon as you work in the derived category. I just find myself unable to pin down why exactly this is the case.
Any ideas anyone?
You seem to suggest that hypercohomology arise only when you have a sheaf $\mathcal{F}$ and a resolution $\mathcal{G}^\bullet$ by non-necessarily acyclic sheaves. Well, sure $\mathbb{H}^i(X,\mathcal{G}^\bullet)$ is hypercohomology, but this is the same thing as $H^i(X,\mathcal{F})$ which is (not hyper) cohomology.
I would say that for hypercohomology, we would rather start with a complex $\mathcal{F}^\bullet$ which may not be quasi-isomorphic to a single object. For example, if $f:\mathcal{F\to G}$ is a morphism which is neither injective nor surjective, then this can be seen as complex concentrated in (homological) degree 1 and 0, and you can consider the hypercohomology $\mathbb{H}^i(X,\mathcal{F}\xrightarrow{f}\mathcal{G})$. This is interesting because it fits into a long exact sequence $$...\to H^i(X,\mathcal{F})\to H^i(X,\mathcal{G})\to\mathbb{H}^i(X,\mathcal{F}\xrightarrow{f}\mathcal{G})\to H^{i+1}(X,\mathcal{F})\to ...$$
In this regard, when you have a complex of sheaves $\mathcal{F}^\bullet$, you don't need (except for some constructions of hypercohomology) to construct a bicomplex $\mathcal{G}^{\bullet,\bullet}$ to take "hyperhypercohomology".
Now you can ask the following question : if I have a bicomplex $\mathcal{F}^{\bullet,\bullet}$ (first quadrant for simplicity), but not necessarily a resolution of a single complex, what would be the natural extension of cohomology/hypercohomology for it ?
Well, as you realized, there is nothing new here : the natural extension would be hypercohomology of the total complex. Indeed :
if $\mathcal{G}^{\bullet,\bullet}$ is column-wise (or row-wise) a resolution of a complex $\mathcal{F}^\bullet$, then $\mathcal{F}^\bullet\to\operatorname{Tot}^\bullet(\mathcal{G})$ is a quasi-isomorphism. So $\mathbb{H}^i(X,\mathcal{F}^\bullet)\simeq \mathbb{H}^i(X,\operatorname{Tot}^\bullet\mathcal{G})$ expanding $H^i(X,\mathcal{F})\simeq\mathbb{H}^i(X,\mathcal{G}^\bullet)$.
working row-wise and then column-wise leads to certain spectral sequences which converges to the total complex. (Well this argument is a bit backward but still shows the relationship between a bicomplex and its total complex).