Back of the envelope time.
First, we have to assume perfect data, so the only factor at play here is whether there is a geometric eclipse or not. Of course if you have poorer data then you will miss some planets because they are too small. i.e. We are looking for the fraction that can be detected in principle.
Let's assume the planets are small enough that their size does not really influence the transit probability, which is given by $\sim R_*/a$, where $a$ is the semi-major axis.
Let's assume circular orbits.
Let's assume a bare planet with no albedo so that the equilibrium temperature is given by
$$ T_{\rm eq} = T_* \sqrt{\frac{R_*}{2a}}\ .$$
Let's assume that a typical star in the Galaxy is an M-dwarf with a temperature $T_* \simeq 3500$ K and a radius of $R_* = 0.5 R_{\odot}$, and let's assume that the likelihood of planet occurrence is independent of stellar mass, so that the properties of an M-dwarf can be assumed for statistical purposes (in practice the answer will depend on what kind of star you are considering).
Let's assume that a habitable planet needs the equilibrium temperatures to be between 273K and 350K (arbitrary I know, and ignores the issue of atmospheres). The range of $a$ for this temperature range, around our fiducial M-dwarf, is between $50R_{\odot}$ and $82R_{\odot}$, with a probability of detecting a transit of between 0.6-1.0%.
So that is my answer 0.6-1%
The probability for higher mass stars is smaller and there are fewer of them. That is because although they are larger, the planets have to be much further away to be in the habitable zone (e.g. the transit probability for Earth is 0.4%). The main uncertainty is the probability of close-in planet occurrence for very low-mass stars where the transit probability can be much higher even though there are fewer host objects.