If $P \implies Q \implies R$ holds, does it follow that $\lnot R \implies \lnot Q \implies \lnot P$ is also true?

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I know that implications are not associative.

So I guess my question is: What interpretation for $$P \implies Q \implies R$$ should I use, so that $$\left(P \implies Q \implies R\right) \equiv \left(\lnot R \implies \lnot Q \implies \lnot P\right)?$$

Does it suffice to use $$\left(P \implies Q \implies R\right) \equiv \Bigg(\left(P \implies Q\right) \land \left(Q \implies R\right)\Bigg)?$$

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You want to be a little careful in not confusing logical notation with mathematical notation.

In particular, the logical operator called the material implication (more often written using $\to$ rather than $\implies$) is indeed not associative, and so $P \to Q \to R$ is ambiguous ... does it mean $(P \to Q) \to R$ or $P \to (Q \to R)$? On neither interpretation will it be equivalent to the corresponding interpretation of $\not R \to \neg P \to \neg Q$ though.

On the other hand, in mathematical notation you may sometimes find an expression like $P \implies Q \implies R$, but that kind of notation is really a short-cut notation to express two logical implications. So here, the $\implies$ is no longer a logic operator used within the language of logic, but rather a meta-logical expression expressing that (given some basic assumptions/axioms of the mathematical domain that we're working in) $Q$ can be derived from $P$, and that $R$ can be derived from $Q$.

But you really do not want to write that as $(P \implies Q) \land (Q \implies R)$, because the $\land$ is a logical operator used for in expressions of formal logical, while something like $P \implies Q$ is not a logical expression (again, it's more of a meta-logical expression).

So, you should just say that $P \implies Q \implies R$ is to be interpreted as $P \implies Q$ and $Q \implies R$. And yes, that also means that $\neg R \implies \neg Q$ and that $\neg Q \implies \neg R$, which using the same short-cut mathematical short-cut notation can be written as $\neg R \implies \neg Q \implies \neg P$