Yes you appear to "understand the whole wavefunction / probability distribution thing right", and it is possible to measure a probability of presence in the forbidden region, but this probability is usually small.
I don't know whether any experiment has shown it with "real particle" like atoms, electrons or neutron, but it has been definitely shown with photons. The "classically forbidden region" in the particle pictures corresponds to evanescent waves, and there have been experiment where evanescent waves have been used to excite the fluorescence of atoms. Each photon emitted by an atom in the evanescent wave can then be seen as a position measurement of a photon in the evanescent wave region.
The idea is the following. You have three zones. One before the barrier, $x<0$, with $V(x)=0$ (zone 1), one inside (or above) the barrier $0\leq x\leq a$ with $V(x)=V_0>0$ (zone 2), and one after the barrier with $V(x)=0$ (zone 3).
Let's say that a particle comes from the left, with energy $0<E<V_0$, so it will go "inside" the barrier. Then, you'll have plain wavefunctions in the zones 1 and 3, where the energy is greater than the potential, and real exponential functions in zone 2, something like
$$\psi _1(x) = Ae^{ik_1x} + Be^{-ik_1x}$$
$$\psi _2(x) = Ce^{k_2x} + De^{-k_2x}$$
$$\psi _3(x) = De^{ik_1x} + Ee^{-ik_1x}$$
This comes from the solution of the time-indepent Schrödinger equation with a constant potential. From your question I assume you know how to reach this point (comment me if you don't). Note that I've used the same $k_1$ in zones 1 and 3 because the potential is the same.
Now you have to apply boundary conditions. Of course, $\psi$ must be continuous and derivable, so we should apply that:
Between zones 1 and 2, in $x=0$
$$\psi_1(0) = \psi_2(0)$$
$$\psi_1'(0) = \psi_2'(0)$$
Between zones 2 and 3, in $x=a$
$$\psi_2(a) = \psi_3(a)$$
$$\psi_2'(a) = \psi_3'(a)$$
Well, you have 6 constants, but only 4 boundaries. Of course, the final goal is to write the final wavefunction $\psi(x)$ in terms of only ONE constant (let's say $A$). To achieve this, I need 5 boundary conditions. But I only have 4. What should I do?
In this case I can eliminate one constant on zone 1 or 3. Whatever constant I want. Do you want to eliminate $B$? Do it. Maybe $D$? Do it then. The problem is not going to change. Try it and convince yourself. But I would choose to eliminate $E$, the wave in zone 3 that comes from right to left. Why? Simply because I imagine a particle coming from left. This will reflect between zones 1 and 2, and also between zones 2 and 3. But in zone 3 I only have transmission: only a particle going from left to right. Although any constant can be eliminated, we eliminate the one we want according to our classical thought. Make that constant 0. Now your 4 boundaries are enough.
Once you have the expression of the final $\psi(x)$ in terms of a constant. This final function will have plain waves in the zone 1 and also in 3, so it won't be normalizable in the usual way:
$$\int _{-\infty}^{+\infty}|\psi(x)|^2 dx = 1$$
because sines and cosines are not square-integrable. If you want to think on normalization, you can think on some alternative ways of normalization, for example: instead of an infinite $X$ axis, you could have very big box where $\psi(L)=\psi(-L)=0$. Then your function would be normalized. Of course, you must think that $L$ is large enough so you can think on a barrier (and not on a barried inside an infinite well). Also you can take a look to "rigged Hilbert spaces" for discussion on this non square-integrable functions.
In any case, you should forget about a normalization with plain waves. The big box concept will do the trick. Instead of that, in this kind of problems you're interested in the reflection and transmission coefficients. They're based on probability currents and this things, but in this simple cases can be calculated as
$$T = \dfrac{|B|^2}{|A|^2}$$
$$R = T - 1$$
Since $B$ is the wave which go left in zone 1 (reflection).
You finally calculate $R$ writting $B$ and $A$ in terms of the constant you've chosen before (if you've chosen $A$, it will be easy :D). That will eliminate all the constants and say you how many particles will be able to go through the barrier and how many will reflect -that is, in fact, the thing you can measure in a real experiment.
Best Answer
Yes.
The solution you mentioned (time-independent decaying exponential) is a solution of the time-independent Schroedinger equation $$ \hat{H}\psi = E\psi $$ Since the solutions of this equation depend on time only via factor $e^{\frac{Et}{i\hbar}}$, the resulting probability density $|\psi|^2$ is time-independent.
To obtain time-dependent probability density and non-zero probability current, you can attempt to solve time-dependent Schroedigner equation $$ \partial_t\psi = \frac{1}{i\hbar}\hat{H}\psi $$ for initial condition $\psi(x,t_0)$ that is no equal to above $\psi$ function, for example some localized wave packet. But this is quite difficult and may require numeric computer calculation, so it rarely occurs in textbooks.