You're extremely confused about what it means to define an operation.
2) How do you compute it?
You tell me! You invented it. The question is like the inventor of chess asking how bishops move - it's your game, dude. Or, to take a more mathematical example, suppose I were to say "I'm going to define a new class of numbers called Pretty numbers. For example, by definition $3$ and $12$ are pretty numbers. Now, can you help me figure out if $20$ is a pretty number?".
This is what a definition of an operation looks like:
Define the operation $\star$ on integers by $a\star b=a^2 +\text{the last digit of b}$.
Now I have a rule that lets me calculate $a\star b$ for any pair of integers $a$, $b$. What you've done is defined $a@a$ for any integer $a$. There is no "correct" answer to what $3@2$ equals, you have to define it. Until you've defined it, the answer doesn't exist.
Furthermore, you're not really "allowed" to just define $b@b@b$. If $@$ is a binary operation, then it has to have two arguments, not three. When you write down $b@b@b$, do you mean $(b@b)@b$, or $b@(b@b)$? What if there is no operation $@$ such that $x@x=x+2$ for all $x$, but also $x@(x@x)=x+3$ for all $x$?
Here's a way of reformulating your question to give it meaning:
Is there any operation $@$ on $\mathbb Z$ such that for all $x$, we have:
$$x@x=x+2$$
$$x@(x@x)=x+3$$
$$x@(x@(x@x))=x+4$$
and so on?
Notice that we're not defining some object and then asking about its properties, we're listing some properties and asking if such an object exists.
Well, let's see. If such an operation existed, then the conditions would imply that for all $x$, and all $n\geq 2$. $x@(x+n)=x+n+1$. This determines $x@y$ whenever $y\geq x+2$: $x@y=x+(y - x)+1=y+1$. In fact, by induction, you can show that as long as an operation verifies $x@y=y+1$ for pairs $(x, y)$ that verify $y\geq x +2$, the required condition will be satisfied, and you can define it in any way you like for other pairs $(x, y)$.
