What is the oxidizing agent for Fe in the following redox reaction?
To answer your initial question:

The oxidizing agent is the reagent that gets reduced. Conversely: The reducing agent is the reagent that gets oxidized.

Now, addressing your trouble with the concept:
You are right that hydrochloric acid is found in the ionic form when in aqueous solution, so:
$$\ce{HCl -&gt;[][\ce{H2O}] H+ + Cl-}$$

Now we have the reagent that gets reduced, which is the proton. Together with another proton and two electrons, they form hydrogen gas that then evolves from the solution:
$$\ce{2H+ + 2e- -&gt; H2 ^}$$
This is basically our electron sink, because that's where they go "to die", i.e. to vanish from our solution.

But where do these electrons come from? Well, we have two potential electron donors here: The chloride anion and the iron. Which one will give up his electrons more easily?

The answer we already know: Iron.

Why? Because as you correctly stated, the chloride anion has all the shells filled and is very happy because of that filled octet state. So the energy to get out an electron from it is higher than to rip it away from an unsuspecting iron atom.1)
$$\ce{Fe^0 -&gt; Fe+ + e- -&gt; Fe^{2+} + 2e-}$$

The electrons then vanish (because of the formation of hydrogen, remember?) which leaves us with some iron cations and chloride anions. They will then start to mix with each other. Don't judge them, they don't know any better!
$$\ce{Fe^{2+} + 2Cl- &lt;=&gt; FeCl2}$$

Since ferric chloride (II) is quite soluble in water, it will most probably not precipitate.



Footnotes:


Actually, you could also compare the standard reduction potentials of the reactions $\ce{Fe -&gt; Fe^{2+} + 2e-}$ and $\ce{Cl- -&gt; Cl^{.} + e-}$. The one that has the lower standard reduction potential is the winner (because the energy of a system is always minimized). This is also what you would usually do, but in this example it wasn't necessary because you could get to the solution via chemical concepts.