**The answer is no in general, but yes if the sequence $Y_n$ is non-negative.**

First, let us focus on the case where $Y_n$ is non-positive. Then, $\sup \frac{1}{n}|Y_n|=-\inf \frac{1}{n}Y_n$. If you assume moreover that all the $Y_n$ are $L^1$, then by Kingman's subadditive theorem, $\frac{1}{n}Y_n$ converges to $Y=\inf \frac{1}{n}Y_n$ almost surely. Notice then that $\mathbb{P}\left (\sup\frac{1}{n} |Y_n|\geq c\right )=\mathbb{P}\left (Y\leq -c\right )$. Whenever $Y$ has positive probability of taking the value $-\infty$, you cannot bound $\mathbb{P}\left (Y\leq -c\right )$ by something converging to 0 as $c$ goes to infinity.

Here is a concrete counter-example. Let $Y_n$ be the constant function $Y_n=-n^2$. Then $Y_n$ is subadditive and satisifes all your assumptions. You have $\sup \frac{1}{n}|Y_n|=+\infty$ and so for any $c$, $\mathbb{P}\left (\sup\frac{1}{n} |Y_n|\geq c\right )=1$, so you don't have $\mathbb{P}\left (\sup\frac{1}{n} |Y_n|\geq c\right )\leq \frac{1}{c}\mathbb{E}(|Y_1|)=\frac{1}{c}$.

However, small remark : the answer is yes for non-positive $Y_n$ if you have the property that $\mathbb{E}(\frac{1}{n}|Y_n|)\leq \mathbb{E}(|Y_1|)$. Indeed, using Markov inequality, you get $\mathbb{P}\left (\frac{1}{n} |Y_n|\geq c\right )\leq \frac{1}{c}\mathbb{E}(\frac{1}{n}|Y_n|)\leq \frac{1}{c}\mathbb{E}(|Y_1|)$ and this is true for all $n$, so this is true for the almost sure limit, using dominated convergence.

About the non-negative case now. Inequality (3) is usually stated without the absolute values in literature :
$$\mathbb{P}\left (\sup\frac{1}{n} Y_n\geq c\right )\leq \frac{1}{c}\mathbb{E}(|Y_1|).$$

**This statement is true in general** and so in particular, the answer to your question is yes whenever $Y_n$ is non-negative.

Indeed, consider a subbaditive sequence $Y_n$, that is satisfying your condition $Y_{n+m}\leq Y_m+Y_n\circ \tau^m$. Let $Z_n=\sup_{k=1,...,n}\frac{1}{k}Y_k$. Also let $\widetilde{Y}_n=\sum_{j=0}^{n-1}Y_1\circ \tau^j$ and finally, let $\widetilde{Z}_n=\sup_{k=1,...,n}\frac{1}{k}\widetilde{Y}_k$. As you claim, the result is true for the sequence $\widetilde{Y}_n$.

Note that since $Z_n$ in non-decreasing, you have $\mathbb{P}\left (\sup\frac{1}{n} Y_n\geq c\right )=\lim_n\mathbb{P}(Z_n\geq c)$ so we just need to prove that $\mathbb{P}(Z_n\geq c)\leq \frac{1}{c}\mathbb{E}(Y_1)$.

Now for fixed $n$, for every $x$, there exists $1\leq k(x)\leq n$ such that $Z_n=\frac{1}{k(x)}Y_{k(x)}$. Because of subadditivity, you have $Y_{k(x)}\leq \sum_{j=0}^{k(x)-1}Y_1\circ \tau^j(x)=\widetilde{Y}_{k(x)}(x)$.
So $\frac{1}{k(x)}Y_{k(x)}\leq \frac{1}{k(x)}\widetilde{Y}_{k(x)}(x)\leq \widetilde{Z}_n(x)$. This proves that for any $x$, $Z_n(x)\leq \widetilde{Z}_n(x)$ so $\mathbb{P}(Z_n\geq c)\leq \mathbb{P}(\widetilde{Z}_n\geq c)$. Using that $\widetilde{Z}_n$ also is non-decreasing, you get $\mathbb{P}(\widetilde{Z}_n\geq c)\leq \mathbb{P}\left (\sup\frac{1}{n} \widetilde{Y}_n\geq c\right )$ and so you can use the result for $\widetilde{Y}_n$.

withthe absolute value) are known to be true when $(Y_n)$ is additive. It's easy to see (by more or less precisely the same proof as in the additive case) that $(2)$ remains to hold true when $(Y_n)$ is subadditive, but it's not clear to me how $(3)$ generalizes and how this generalization can be proven. $\endgroup$ – 0xbadf00d Apr 30 '20 at 14:111more comment