Answer: $f(x) = a - x$ for arbitrary real constant $a$.

First set $y = 0$ in the initial inequality
$$f(x + y) + y \le f(f(f(x))) \quad (*)$$
thus obtaining
$$f(x) \leq f(f(f(x))) \quad \forall x \in \mathbb{R}. \qquad (1)$$
Further, set $y = f(f(x)) - x$ in $(*)$. Then
$$f(f(x)) \leq x \quad \forall x \in \mathbb{R}. \qquad (2)$$
Replacing $x$ by $f(x)$ in (2) we obtain $f(f(f(x))) \leq f(x)$ which together with (1) gives $f(f(f(x))) = f(x)$.

Now $(*)$ becomes
$$f(x + y) + y \leq f(x) \quad \forall x, y \in \mathbb{R}. \qquad (3)$$
Set $x = 0$ in (3), then
$$f(y) \leq a - y \quad \forall y \in \mathbb{R}, \qquad (4)$$
where $a = f(0)$.
Finally, set $y = -x$ in (3) thus getting
$$a - x \leq f(x). \qquad (5)$$
Comparing (4) and (5) we obtain $f(x) = a - x$.

It is easy to verify that the function $f(x) = a - x$ satisfies the given inequality for any real number $a$.