How do we know $\neg \neg LEM$ isn't provable in MLTT?
The task here is indeed to find a model of MLTT in which $\neg LEM$ holds (and so $\neg\neg\neg LEM$ holds as well). Realizability models have this feature, for instance; see also this. Here, MLTT functions are interpreted with codes of computable recursive functions, so the usual uncomputable functions, e.g. halting oracles, cause $LEM$ to fail.

If you define $LEM$ as ranging over all types, and not just types which are propositional (i.e. have at most one inhabitant up to propositional equality), then it is also the case that $LEM$ is inconsistent with the univalence axiom of homotopy type theory, so this version of $LEM$ is also refuted by any model which can interpret the univalence axiom. Such are the simplicial set and cubical set models. 

$LEM$ is also inconsistent with parametricity, so models validating parametricity also refute $LEM$. I think formally these are considerably simpler than univalent or realizability models; see for example the reflexive graph model. Intuitively, this works by noticing that MLTT does not allow inspecting the structure of types, so any function with type $\prod_{A : Type} A \rightarrow A$ must be the identity function. However, $LEM$ allows one to branch on whether two types are equal, allowing a function with the above type which is not the identity function on all types, e.g. negates Boolean inputs.