Why is diffraction related to wavelength, not amplitude? For diffraction, the wavelength of the incident beam should be in range magnitude of the slit length, but why the isn't the amplitude related to the length of the slit?
 A: remember that the amplitude is not an amplitude in space, it is an amplitude in the sense of the intensity of the electromagnetic field. The spatial amplitude is given by the wavelenght
A: The slit is a source of the wave. The wave equation solution (diffraction) depends on the slit size and on the wave wave-length because each point of the slit is an elementary source of a spherical wave. They all add up and produce the resulting diffraction pattern.
The wave amplitude determines the overall intensity, not the diffraction angles.
A: Phases are determined by wavelengths not magnitude of amplitudes. Phases between waves are crucial at obtaining interference and diffraction pattern.
A: Who says that the amplitude is not related to the slit length? I understand that you ask of the amplitude of the diffracted beam.
Note : you use a not very good expression, slit-length. You have to say slit width, because in some cases the slit has a very, very big length and a width of the order of $\lambda$, in other cases the slit is circular. The formula that I indicated is for the very long slit and width of the order of $\lambda$.
So I use the expression slit-width not slit-length.
Now, here is the formula calculated in the Fraunhofer regime i.e. far from the single-slit
$I(\theta) = I_0 \ sinc^2(\frac {d\pi}{\lambda} sin(\theta))$,
where $I(\theta)$ is the intensity of the pattern in the direction $\theta$, $d$ is the slit-width, and $\theta$ the angle under which a certain point in the pattern is viewed from the center of the slit. The intensity is the absolute square of the amplitude (of which you ask).
You can see at the site that I indicated that if the slit is much wider than $\lambda$ one sees clearly the central maximum, the other maxima are very weak. So, the slit width, more exactly the ratio $d/\lambda$, influences the clarity of the pattern. But what decides where are the minima and maxima is mainly $sin(\theta)$. As to $I_0$ it influences the overall luminosity of the pattern.
A: I'll rephrase the question as I interpret it:

Electromagnetic waves are drawn like this:
 source:
  https://commons.wikimedia.org/wiki/File:Onde_electromagnetique.svg
Suppose this wave comes up to a vertical slit (a slit in the
  z-direction). What if the red arrows are longer than the slit? Then
  the wave won't fit through. But if the arrows are smaller than the
  slit, the wave will fit through. Shouldn't we see different behavior
  depending on whether the arrows in the picture fit through the slit?

The answer is no, because the picture is deceptive. The arrows don't have a length in terms of centimeters. The entire arrow is completely located along the x-axis in this picture. There is just no way to draw the arrow so that the entire thing is located at a point, but that's what you have to imagine. If we made the electric field half as strong, the arrows would be half as long, but all that means is that a charge at that location would feel half as much force. It doesn't mean the arrows would point to different physical points in space.
At other locations in space, there would be different arrows, but those aren't drawn either, for simplicity in the diagram. You can play around with something like https://phet.colorado.edu/en/simulation/radio-waves to try to visualize this a bit more.
The red arrows in the picture have dimensions of electric field strength - Volts/meter. You can't compare the amplitude of the electric field to the size of the slit for the same reason you can't compare a speed to a mass; they simply have completely different dimensions.
