Is there a magnetic attraction between two parallel electron beams? I am refering to Ampere's force law, and to the beams accelerated after the cathode, so the deflection is not due to their respective cathode. In other words, do two electrons accelerating parallel to each other converge because of magnetic attraction?
Does it apply to wires only or to a beam of charge carriers too?
If coulomb interaction is stronger and make them diverge, is there at least a small attraction limiting it?
 A: If you look at the problem in the electrons' stationary reference frame, where there is only a Coulomb force and no magnetic field, then you can see that there is always repulsion between the electron beams.
All observers in other reference frames will agree that there is a repulsive force between the electron beams, but in the stationary lab frame, the electric field in the stationary electron frame will be transformed into an electric field plus a magnetic field, as expected because the moving electrons constitute a current.
The total force on an electron will be given by the Lorentz formula 
$$\vec{F} = -e\vec{E} -e\vec{v}\times\vec{B},$$
where $v$ is the electron velocity in the lab frame.
The two terms in this expression are in opposite vector directions, so the magnetic term acts to reduce the repulsion between the electron beams. However, it can never be large enough to overcome the Coulomb repulsion. To see this, we note that if the E-field in the stationary frame of the electrons, at the position of an electron in one of the beams, is $\vec{E'}$, then to an observer in the lab, moving with a relative velocity $\vec{v}$, the electric and magnetic fields at this position in the lab frame are:
$$ \vec{E} = \gamma \vec{E'}$$
$$ \vec{B} = \frac{\gamma}{c^2} \vec{v} \times \vec{E'},$$
where $\gamma = (1 - v^2/c^2)^{-1/2}$.
In the stationary electron frame, the Lorentz force on the electron is
$$ \vec{F'} = -e\vec{E'}\ .$$
In the lab frame it is
$$ \vec{F} = -e\gamma \vec{E'} - e\frac{\gamma}{c^2}\vec{v}\times (\vec{v}\times \vec{E'})\ .$$
Because $\vec{E'}$ and $\vec{v}$ are perpendicular and $\vec{a}\times (\vec{a}\times \vec{b}) = (\vec{a}\cdot \vec{b})\vec{a} - a^2 \vec{b}$, then
$$\vec{F} = -e\gamma \vec{E'} + e\gamma \frac{v^2}{c^2} \vec{E'}$$
$$ \vec{F} = -e \vec{E'} \gamma (1 - v^2/c^2) = \frac{F'}{\gamma}$$
Since $\gamma \geq 1$, then the perpendicular force between the electrons is reduced (by a factor of $\gamma$), but it cannot change sign.  
A: In the lab frame there is a magnetic attraction, but it will never overpower the Coulomb repulsion between the two beams.
This is easiest to see in a frame of reference which moves with the electrons themselves: there, the electrons are stationary, and the only force between them is the repulsive Coulomb force. That said, if the electrons are moving fast enough (and, since the problem is scale-free, any velocity is "fast enough"), special relativity will require some minor tweaks to how that repulsion is observed from the lab frame, because of effects coming from length contraction and time dilation. 
In the lab frame, those relativistic corrections to the Coulomb repulsion can be interpreted as an additional force which is proportional to the velocities and to the charge of the electrons. This is what we know as the magnetic interaction between the two beams.
If you want to see this line of understanding in all its glory, I recommend the relativity-and-magnetism chapter ('The fields of moving charges') in Ed Purcell's Electricity and Magnetism. 
