The probability of finding a particle at a point is always zero.
Recall that $\rho(x) = \lvert\psi(x)\rvert^2$ is a probability density, not a probability, and so the probability to find the particle somewhere inside the interval $[a,b]$ is given by
$$ P([a,b]) = \int_a^b \rho(x)\mathrm{d}x.$$
Since points have zero measure ($\int_a^a \rho(x)\mathrm{d}x = 0$ regardless of $a$), this is always zero for single points. So it is not evident that there is any meaning to saying "the particle will never be found at $x_0$" because quantum mechanics only allows us to talk meaningfully about a region (however small) in which the particle can be found.
This is supported by the fact that the "eigenstates" $\lvert x\rangle$ of the position operator are not actual states since they are non-normalizable ($\langle x \vert x\rangle$ cannot be made finite/well-defined), so there is no actual measurement whose result could be $\lvert x\rangle$, a fully localized particle.
However, you are asking about the nodes of the wavefunction of a particle trapped in a box. Indeed, even though you should not think of them are "points where the particle can never be found", the $n$-th excited state has $n$ of these nodes in its wavefunction.
garyp suggests an alternative interpretation of "the particle can never be found at $x_0$" in the comments that actually then is correct for the nodes:
For the nodes $n_i$, we have
$$\lim_{a\to 0}\frac{P([n_i-a,n_i+a])}{P([x_0-a,x_0+a])} = 0$$
for any $x_0$ that isn't a node itself. This means, in words, that if we take regions of equal size centered around the points $n_i$ and $x_0$ and shrink them, it becomes more and more likely to find the particle around $x_0$ compared to finding it around $n_i$, until in the limit the ratio becomes zero suggesting it is infinitely more likely to find the particle "at" any other $x_0$ then it is to find it "at" $n_i$. Note that the latter part of this sentence should only be understood heuristically due to the actual probability of finding a particle at a point being zero as discussed at the beginning.