I've been on constant search for any proof regarding the Law of Triple Negation. I happened to find Brouwer's paper, and he provided a theorem which seemed to be the law of triple negation: $$ \lnot P \leftrightarrow \lnot \lnot \lnot P $$
How would one go about translating the theorem and its proof into modern terminology?
You are right about what Brouwer is asserting. Here is how I would prove it.
1) Assume $P \to Q$ and $((P \to Q) \to Q)$, then you can conclude $Q$. Discharging the assumptions, you have proved $(P \to Q) \to (((P \to Q) \to Q) \to Q)$.
2) Assume $P$ and $P \to Q$, then you can conclude $Q$. Discharging the assumptions, you have proved $P \to ((P \to Q) \to Q)$.
3) Assume $((P \to Q) \to Q) \to Q$ and $P$. By (2) and the second assumption, you have $(P \to Q) \to Q$ and then by the first assumption you have $Q$. Discharging the assumptions, you have proved $(((P \to Q) \to Q) \to Q) \to (P \to Q)$.
4) By (1) and (3) you have $(P \to Q) \leftrightarrow (((P \to Q) \to Q) \to Q)$. Put $Q = \bot$ (or, if you prefer, $\mathsf{false}$) so that $A \to Q$ is equivalent to $\lnot A$. This gives $\lnot P \leftrightarrow \lnot\lnot\lnot P$.
Note that this doesn't use any special properties of $\bot$. It doesn't require ex falso quodlibet.
Here's a comparison with Brouwer's proof as given in your link:
Brouwer's "Firstly" is my (3) which Brouwer proves by observing that $(P \to Q) \to (\lnot Q \to \lnot P)$ and uses the "established fact" that $P \to \lnot\lnot P$ to conclude that $\lnot\lnot\lnot P \to \lnot P$.
Brouwer's "Secondly" is my (1), which Brouwer proves by instantiating $Q$ in $Q \to \lnot\lnot Q$ as $\lnot P$ to conclude $\lnot P \to \lnot\lnot\lnot P$.
As a light-hearted aside: Either Brouwer or the translator of the text in your link is to be "congratulated" ($\ddot{\frown}$) for a fine way to make a proof of an equivalence incomprehensible: "Firstly, blah, blah, blah, so we conclude the implication of $A$ by $B$. Secondly, dudum, dudum, dudum, so $A$ implies $B$.