In the comments, you ask
Are strongly compact cardinals superstrong?
The answer is no, not necessarily, and indeed, strongly compact cardinals need not even be strong, nor even a little bit strong. The reason is that it is relatively consistent that the least strongly compact cardinal is the same as the least measurable cardinal, and in this case such a cardinal $\kappa$ will not even be $(\kappa+2)$-strong. This was the first instance of the so-called "identity crises" phenomenon, discovered by Magidor, and similar phenomenon have now been uncovered for many other cardinals (for example, the least weakly compact cardinal can be unfoldable, weakly measurable and nearly $\theta$-supercompact).
That is for outright implication, but meanwhile, for consistency strength, it is a different story. As far as we know, strongly compact and supercompact cardinals have the same consistency strength. Here, the relevant observation is:
Theorem. If $\kappa$ is $2^\kappa$-supercompact, witnessed by $j:V\to M$, then $\kappa$ is superstrong in $M$.
This is the reason that supercompact cardinals are strictly stronger than superstrong cardinals in consistency strength.
One can prove the theorem by noting that the induced extender determined byProof: The map $j\upharpoonright P(\kappa)$$j\upharpoonright V_{\kappa+1}:V_{\kappa+1}\to M_{j(\kappa)+1}$ has size $2^\kappa$ and is therefore inside $M$. Using this part of $j$ to define an extender $E$, the derived extender of $j$, the model $M$ can produce an embedding $j_E:M\to N$, which agrees with $j$ on $V_{\kappa+1}$. In particular, this means $M_{j(\kappa)+1}\subset N$, which means that $E$ witnesses that $\kappa$ is superstrong from the perpsective of $M$. QED
Perhaps one of the inner model theorists will post an answer explaining how much of the inner model theory one can undertake from a strongly compact cardinal, and I suppose that is really what you are asking.