'size' of a photon What's the smallest aperture a photon can pass through? I mean with no transmission at all.
I'm pretty sure I saw long ago an experiment when they were reducing the size of a hole in a gold film and at about a quarter or a third of the wavelength there was no transmission. 
If so, can we say that the size of the aperture can be associated with the dimension of the photon?
 A: I should like to capture the excellent comment by user CuriousOne (emphasis mine):

There is no such size. A smaller aperture will merely reduce the transmission probability, but there is no known cutoff. Indeed, there is currently a lot of interest in deep sub-wavelength imaging techniques. A photon, by the way, is not an object. It's a quantum, i.e. the discrete unit of change of a quantum field. It does not have a size any more than the color red does.

So the only precise version of the OP's question is "what is the transmission probability through an aperture" and the answer to that question is that once the aperture is subwavelength, then the transmission probability drops off exponentially with the thickness of the screen the aperture pierces. As CuriousOne says, techniques such as scanning nearfield optical microscopy (SNOM) can image arbitrarily small features by gathering light through abitrarily small apertures: the catch is the total thickness represented by the aperture "screen" and standoff must be much less than the size of feature imaged. The essential problem is that electromagnetic fields couple through subwavelength holes through evanescent wave coupling. If the screen is of thickness $t$ and the size of the aperture $d$, the transmission probability is estimated by:
$$\begin{array}{lcl}p(t,\,d) &\propto &\exp\left(-4\,\pi\,z\,\sqrt{\frac{1}{d^2}-\frac{1}{\lambda^2}}\right)\\&\approx&\exp\left(-4\,\pi\,\frac{z}{d}\right);\quad d\ll\lambda\end{array}$$
This is a horrifically fast dropoff. If your screen thickness is$1{\rm \mu m}$ and the aperture $50{\rm nm}\ll\lambda$, the probability is of the order of $10^{-110}$.
See my answer to question a related idea: the limits on subwavelength imaging with light for further information.
A: Light is made up of photons which are elementary particles. The standard model of particle physics fits  the data very well with the  hypothesis that all elementary particles are point particles.
In an experiment where the size of a slit is varied and the cross-section for a photon to pass through the slit is measured two consideration should enter when comparing the probability of a photon to go through the slit:
$\bullet$ The wavelength of the photon which defines its momentum $\dfrac{h}{\lambda}$,
$\bullet$ The size of the slit .
This, before solving the quantum mechanical problem "photon +slit"   scattering, has the rule of thumb of the Heisenberg Uncertainty principle :
$$\begin{align}\color{red}{\Delta \mathbf{x}\Delta \mathbf{p} \gt \frac{\hbar}{2}}\\ \color{red}{\Delta \mathbf{E}\Delta \mathbf{t} \gt \frac{\hbar}{2}}\end{align}$$

I'm pretty sure I saw long ago an experiment when they were reducing the size of a hole in a gold film and at about a quarter or a third of the wavelength there were no transmission. 

Without a link it is hard to accept this as a fact. A quarter or a third of the HUP constraints would allow a quarter or a third of the incoming beam photons to go through, if one ignores the special surface of gold. If it is a fact that means that one needs a solution of the quantum mechanical set-up " photon + gold surface +hole " scattering, which will increase the probability of the photon to interact with the fields of the gold surface. Googling I found this, which is the way to approach the problem.  
The photon as an elementary particle is  a quantum mechanical entity, i.e. depending on the boundary conditions its location is either  classical particle like or expressed as a wave like probability amplitude. If your memory is correct the solution is that in this experiment one is observing the probability-wave-like nature  which should be very small for a single photon to pass the hole.
A: If we use the model of an electromagnetic wave, then like theses, a wave with a wave length of lambda need at least a hole with lambda / 2 to escape a cavity. So, the size of a photon could be proposed for a deeper study and experiences as the wave length of the photon divide by 2. But.... just as clue to start ! In fact, what is a photon ! A smily ? ...
A: The simplest answer would be 1 ℓP, Planck length. 
Less than that means no aperture available. 
Unfortunately, it is difficult to imagine what an "aperture" is at that level - a "hole" "through" "what".... 
It might be thought of as a single spatial transaction across (or along) time. Bearing in mind, spatial is not really "spatial", it is planar. 
