Newtons third law with two charged particles Imagine we have two charged particles, $q$ and $Q$.
$q$ is at rest at a point and $Q$ is moving with a velocity. Now $q$ is exerting an electrostatic force on $Q$ and $Q$ makes a magnetic field but because $q$ is at rest there is no force from $Q$ to $q$. 
So, is the third law of Newton violated in this case?
 A: If $q$ has been at rest for a long time, and $Q$ has been moving with a constant velocity for a long time, then the forces they feel from each other are not equal and opposite. But not at all for the reason you stated.
The one that has been stationary at $\vec a$ (for a long time so located at $\vec a(t)=\vec a$) does contribute a constant regular inverse square law electric field like: $$ \vec E(\vec r,t)=\frac{q}{4\pi\epsilon_0}\frac{\vec r-\vec a}{|\vec r-\vec a|^3}.$$
But a charge moving with a constant velocity (for a long time, so located at $\vec b(t)=\vec b(0)+\vec vt$) also produces an electric field and a different one than a stationary charge, the electric field due to $Q$ is a field like: $$\vec E(\vec r,t)=\frac{Q}{4\pi\epsilon_0}\frac{1-v^2/c^2}{(1-v^2\sin^2\theta/c^2)^{3/2}}\frac{\vec r-\vec b(t)}{|\vec r-\vec b(t)|^3}.$$
These are different fields. And when you multiply by the corresponding $q$ and $Q$ they are also different forces. So of course Newton's third law is violated, because charges don't exert forces on each other.
Charges exchange energy and momentum with the electromagnetic field. The energy and momentum of the electromagnetic field is stored as a density in various locations and so can move from where one charge deposited it, through empty space, and then when it gets to where another charge is located it can deliver it.
It's like sending cash in the mail. The amount of cash (energy, momentum) that the sender and receiver have is not conserved. Because sometimes the cash is en route. If you include the cash (energy, momentum) that messengers have then cash (energy, momentum) can be conserved.
And Newton's third law is really about momentum conservation. And momentum is only conserved when you include the field momentum. So you are wrong to think that magnetic fields are why the forces are different. The forces are different, but that's because the electric fields are different.

wouldn't there be force from the magnetic field due to the moving charge $Q\;?$

Absolutely no force from the magnetic field. The stationary charge doesn't produce one for the moving charge to feel, and the and stationary charge is stationary charge doesn't feel a force from the magnetic field of the moving charge. So neither particle feels a magnetic force.
Magnetic fields are important. They are essential for the transportation of momentum from one place to another. And they are essential for correctly transmitting momentum to/from a moving charge.

You emphasized on their position being the same function for a long time? What would happen if the time interval is not long?

If the velocity isn't constant, then firstly there can be additional forces, even self forces. And the electromagnetic fields will be different. In fact the fields caused by acceleration actually are stronger than the other fields when you get far enough away. So you want the time interval to be long enough so that any messages sent between the two of them at speed $c$ would have each of them only getting messages from the other one during a time when it was moving at a constant velocity. Otherwise you need to base the electromagnetic field on the position of the charge back when it sent a message at speed $c$ that just got to you now. And another electromagnetic field based on the velocity of the charge back when it sent a message at speed $c$ that just got to you now. And yet another electromagnetic field based on the acceleration of the charge back when it sent a message at speed $c$ that just got to you now. And even with those there might be additional forces, or even self forces.
When the acceleration has been zero for a long time, that acceleration field is zero where the other charge is now. And the field I wrote is the combination of the position and the velocity field which simplify a lot in the case of constant velocity over a long time because the combination of the fields happens to be related to the position now because the fields due to the position back when the message was sent and the fields due to the velocity back when the message was sent combine to have a field that has a simple relationship to where the charge is now. 
A: $Q $ doesn't generate just a magnetic Field, but also an electric field: Just because $Q $ moves, that doesn't mean that it doesn't generate a field, this field just has to satisfy the Maxwell equations. $\mathbf{\nabla\cdot E} = \rho$ still has to hold, and also the magnetic field that is created is time dependent, so $\mathbf {\nabla\times E} = \dot{\bf B}\;.$ This field creates a force on the resting charge $ q.$
