Terminal velocity is the (asymptotic) maximum velocity that you can reach during free-fall. If you imagine yourself falling in gravity, and ignore air resistance, you would fall with acceleration $g$, and your velocity would grow unbounded (well, until special relativity takes over). This effect is independent of your mass, since
$F = ma = mg \Rightarrow a = g$
Where terminal velocity arises is that air resistance is a velocity-dependent force acting against your free fall. If we had, for example, a drag force of $F_D=KAv^2$ ($K$ is just a constant to make all the units work out and depends on the properties of the fluid you're falling through, and $A$ is your cross-sectional area perpendicular to the direction of motion) then the terminal velocity is the velocity at which the forces cancel (i.e., no more acceleration, so the velocity becomes constant):
$F = 0 = mg - KAv_t^2 \Rightarrow v_t=\sqrt{mg/KA}$
So we see that a more massive object can in fact have a larger terminal velocity.
It's not actually friction, it's the drag force, also called air resistance. Technically, they're different (even though people do sometimes say "air friction"). The drag force depends on the object's speed, shape, and size, as well as the density of the air, but in the ideal case you can assume that the object's shape and size and the density are constant.
In that case, when an object is falling at terminal velocity, the forces on it balance out, which means it has zero acceleration - in other words, its velocity doesn't change. And if the velocity doesn't change, then the drag force, which depends on velocity, won't change either. So the object is "stuck" in a steady state in which the drag force on it will remain constant as it falls.
In practice, though, an object never actually reaches terminal velocity unless it starts at terminal velocity. As it gets closer and closer to terminal velocity, the drag force on it gets closer and closer to exactly balancing out the gravitational force, which means that the net force on it, and thus its acceleration, becomes less and less. Basically, as the object gets closer to terminal velocity, the rate at which it continues to approach terminal velocity gets slower. It'll get continually closer to terminal velocity as time goes on, but it never quite reaches it.
Last year I wrote a blog post about this issue which discusses it using the actual math. It might be of interest to you.
Best Answer
From the equilibrium between drag and weight: $$ \frac{1}{2} C_x \rho v^2 S = m g $$ we can write the terminal velocity as $$ v = \sqrt{\frac{2 m g}{C_x \rho S}} $$ where $m$ is the mass of the phone, $C_x$ its drag coefficient, $S$ its section, $g$ the acceleration of gravity, and $\rho$ the density of air.
Now, this is not really a good answer, as the big question is how to estimate $S$ (depends on the phone orientation) and $C_x$. But at least you can compute an order of magnitude, as in most cases $C_x$ of of order 1.