I really like this question! I can't yet upvote, so I'll offer an answer instead. This is only a partial answer, as I don't fully understand this material myself.
Suppose that $f$ is a quadratic polynomial. Suppose that there is an integer $n$ and a domain $U \subset \mathbb{C}$ so that the $n$-th iterate, $f^n$, restricted to $U$ is a "quadratic-like map." Then we'll call $f$ renormalizable. (See Chapter 7 of McMullen's book "Complex dynamics and renormalization" for more precise definitions.) Now renormalization preserves the property of having a connected Julia set. Also the parameter space of "quadratic-like maps" is basically a copy of $\mathbb{C}$.
So, fix a quadratic polynomial $f$ and suppose that it renormalizes. Then, in the generic situation, all $g$ close to $f$ also renormalize using the same $n$ and almost the same $U$. This gives a map from a small region about $f$ to the space of quadratic-like maps. This gives a partial map from the small region to the Mandelbrot set and so explains the "local" self-similarity.
To sum up: all of the quadratic polynomials in a baby Mandelbrot set renormalize and all renormalize in essentially the same way. (I believe that there are issues as you approach the place where the baby is attached to the parent.) Thus renormalization explains why the baby Mandelbrot set appears.
I always think a discussion of the Mandelbrot set should go along with a discussion of the logistic family, as a simple (the simplest?) interesting model of population dynamics. Of course the logistic family is only the real part of the Mandelbrot set, and here is a first simple pre-calculus exercise: show how to change coordinates to get one parametrization from the other.
(Or: Even show that you can change coordinates for every complex quadratic polynomial to get one of the form $z^2+c$.)
The fact that the Mandelbrot set is bounded was already mentioned in a previous answer. It's easy and straightforward, and well worth covering.
Some other easy exercises would be to determine the fixed points of the polynomials, and the region where there is an attracting fixed point (derivative of modulus less than one), i.e. the "main cardioid" of the Mandelbrot set. Do the same thing for period 2, and maybe ask the question what happens with higher periods - see also comments about density of hyperbolicity below.
It is also possible to discuss the structure of the Julia set (phase space), and in particular the difference between disconnected Julia sets outside M and connected Julia sets inside. While a formal proof would be difficult at this level, giving the geometric idea is not too hard, and for very negative real c (i.e. large $\lambda$ in the logistic parametrization), the result that the invariant set is a Cantor set is easy to do with elementary means (this is done e.g. in Devaney's book "A first course on chaotic dynamical systems").
You say that connectivity of the Mandelbrot set might not be interesting to them, but it's always possible to tell the amusing story about how Mandelbrot's first computer pictures suggested that M is disconnected, but the editor of the paper carefully removed all the little islands from his picture, thinking they were dirt! You could accompany it by running two different algorithms (one which just does a pixel-by-pixel calculation and colors points white or black, making M look disconnected, and e.g. the more standard colored pictures that show the connectedness of level lines quite clearly), and thus making a point about the perils of computer experiments and the importance of mathematical proof (if you care about this).
Finally, I would say that it is a good idea to talk about density of hyperbolicity. The question whether every interior component of the Mandelbrot set corresponds to maps with an attracting cycle is one of the most important open questions in complex dynamics, and yet is quite easy to understand with just a little bit of experimentation. It is always good to show students that even seemingly innocent questions can be the subject of very difficult mathematical research. Of course this is also a chance to mention that density of hyperbolicity in the real case (i.e. density of period windows in the bifurcation diagram) was only established in the 90s, and was a major mathematical breakthrough.
I realize there is a lot here, and it may go beyond what you were looking for, so pick and choose!
Best Answer
Let $S_1$ be a circle of radius $100$ in the complex plane centered at the origin. Clearly everything outside of $S_1$ diverges to $\infty$ under iteration of $P_c$, so the filled Julia set lies entirely inside of $S_1$.
Now take the preimage $S_2$ of $S_1$ under $P_c$. This will be a smaller curve (close to a circle, with a radius of approximately $\sqrt{100} = 10$), which again must contain the entire filled Julia set. Iterating this process, we obtain a sequence $S_1,S_2,\ldots$ of closed curves, each of which contains the filled Julia set in its interior. In fact, since any point outside of the filled Julia set goes to $\infty$ under iteration of $P_c$, the intersection of the interiors of the curves $S_n$ is precisely the filled Julia set. (See this picture for an example. The curves separating the different shades of orange are the iterated preimages of some large circle.)
Unfortunately, this reasoning is not quite correct, because the preimage of a closed curve under $P_c$ is not always a single closed curve. Sometimes it is one closed curve, and sometimes it is two closed curves with disjoint interiors. If we repeatedly take the preimages of $S_1$, we may find that $S_n$ is a union of a very large number of closed curves! Note, however, that these curves still contain the filled Julia set entirely inside of them.
Now here is the key bit: the preimage of a curve will have one component if and only if $0$ lies in the the interior of the preimage. This is because $0$ is the critical point of the map $P_c$. Thus the preimage of a curve is either a single curve that surrounds $0$, or two curves, neither of which surrounds $0$. (In the latter case, the two curves are actually negatives of one another, i.e. symmetric across the origin.) Therefore, there are exactly two cases:
The point $0$ lies in the filled Julia set. In this case, each preimage $S_n$ must be a single curve, so the intersection of the interiors is connected.
The point $0$ lies outside the filled Julia set. In this case, some preimage $S_n$ does not encircle $0$, so it must have two components. Then each successive preimage will have twice as many curves as the last, and the resulting filled Julia set is homeomorphic to the Cantor set. (See this picture for an example. The curves separating the different shades of blue are the iterated preimages of some large circle. You ought to be able to see the first step at which the curve separates into two components.)