Is there actually a 0 probability of finding an electron in an orbital node?
The probability of finding the electron in some volume $V$ is given by:
$$ P = \int_V \psi^*\psi\,dV \tag{1} $$
That is we construct the function called the probability density:
$$ F(\mathbf x, t) = \psi^*\psi $$
and integrate it over our volume $V$, where as the notation suggests the probability density is generally a function of position and sometimes also of time.
There are two ways the probability $P$ can turn out to be zero:
$F(\mathbf x, t)$ is zero everywhere in the volume $V$ - note that we can't get positive-negative cancellation as $F$ is a square and is everywhere $\ge 0$.
we take the volume $V$ to zero i.e. as for the probability of finding the particle at a point
Now back to your question.
The node is a point or a surface (depending on the type of node) so the volume of the region where $\psi = 0$ is zero. That means in our equation (1) we need to put $V=0$ and we get $P=0$ so the probability of finding the electron at the node is zero. But (and I suspect this is the point of your question) this is a trivial result because if $V=0$ we always end up with $P=0$ and there isn't any special physical significance to our result.
Suppose instead we take some small but non-zero volume $V$ centred around a node. Somewhere in our volume the probability density function will inevitably be non-zero because it's only zero at a point or nodal plane, and that means when we integrate we will always get a non-zero result. So the probability of finding the electron near a node is always greater than zero even if we take near to mean a tiny, tiny distance.
So the statement the probability of finding the electron at a node is zero is either vacuous or false depending on whether you interpret it to mean precisely at a node or approximately at a node.
But I suspect most physicists would regard this as a somewhat silly discussion because we would generally mean that the probability of finding the elecron at a node or nodal surface is nebligably small compared to the probability of finding it elsewhere in the atom.
You're quite right: the probability of finding the electron at any single point (or on any particular surface) is zero. Nevertheless, the statement makes sense: what it really means is roughly the following.
Consider a box $V$ with width/depth/height $(w,d,h)$. If all of these are sufficiently small that the wavefunction doesn't variate substantially throughout the box, you can approximate $$ P = \int_V\!\mathrm{d}V\, |\psi(\mathbf r)|^2 \approx |V| \cdot |\psi(\mathbf r_c)|^2 = w\cdot d\cdot h \cdot |\psi(\mathbf r_c)^2| $$ where $\psi(\mathbf r_c)$ is the wave function evaluated at (say) the centre of the box. Now, if that point happens to lie within a nodal plane then the above approximation yields zero.
Strictly speaking, this is of course just wrong: basically the approximation breaks down since there is no zeroth term in the Taylor expansion, hence the dominant term becomes the linear one even on an arbitrarily small range. However, in that more suitable approximation you'd still get “virtually zero” as the result: say the nodal plane is in xy direction and $h$ measures in z direction. Then the integral becomes $$\begin{aligned} P \approx&\, w\cdot d\cdot \int\limits_{-h/2}^{h/2}\!\!\mathrm{d}z\,|\psi(\mathbf r_c + z\cdot \mathbf{e}_\mathrm{z})|^2 \\ \approx&\, w\cdot d\cdot \int\limits_{-h/2}^{h/2}\!\!\mathrm{d}z\,\left|z\cdot\frac{\partial\psi}{\partial z}\Bigr|_{\mathbf{r}_c}\right|^2 \\ =&\, w\cdot d\cdot \left|\frac{\partial\psi}{\partial z}\Bigr|_{\mathbf{r}_c}\right|^2 \cdot \int\limits_{-h/2}^{h/2}\!\!\mathrm{d}z\,(z^2) \\ =&\, w\cdot d\cdot \left|\frac{\partial\psi}{\partial z}\Bigr|_{\mathbf{r}_c}\right|^2 \cdot \frac23 \left(\frac{h}2\right)^3 \propto h^3 \end{aligned}$$ so as you make $h$ small, the probability diminishes not just proportionally as the volume does, but with the third power, making it very small indeed.