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.
Suppose your object is a sphere with a radius $r$ and mass $m$. The aerodynamic drag on a sphere is given by:
$$ F_{drag} = \tfrac{1}{2}C_d \rho \,\pi r^2 \,v^2 \tag{1} $$
where $\rho$ is the density of the air and $C_d$ is the drag coefficient. The drag coefficient varies with speed (the NASA article I linked shows how $C_d$ changes with speed) but over a limited range of speeds it can usefully be taken as constant.
The downward force on the object is simply:
$$ F_{grav} = mg \tag{2} $$
and terminal velocity is reached when the two forces are in balance i.e. when $F_{drag} = F_{grav}$. If we equate equations (1) and (2) we get:
$$ \tfrac{1}{2}C_d \rho \,\pi r^2 \,v^2 = mg $$
and rearranging gives:
$$ v_{term} = \sqrt{\frac{2mg}{C_d \rho\pi r^2}} $$
In your case you keep the size of the spheres constant, in which case we get:
$$ v_{term} \propto \sqrt{m} $$
So terminal velocity does increase with mass. The heavier sphere will have a higher terminal velocity.
Best Answer
Okay, so here's the basic physics: it comes down to differential equations.
Big objects in air tend to see $v^2$ drag. Plugging this into $F = m a$, this means that something which is falling straight downward with speed $v$ will obey the nonlinear differential equation: $$\frac {dv}{dt} = \frac 1\lambda v^2 - g$$for some length $\lambda$.
The terminal velocity is the one where $dv/dt = 0$ hence $v_t^2 = \lambda ~ g,$ giving a simple way to compute $\lambda$ if you know the terminal velocity and the graviational acceleration
Here's the part that you're not going to like: actually solving this equation gives$$\int ~\frac{dv}{v_t}~ \frac{-1}{1 - (v/v_t)^2} = \frac {v_t} \lambda ~ t + C~.$$Then, choosing $v = v_t~\tanh{u}$ we find$$-u = \frac t\tau + C; ~~~v = -v_t~\tanh~\frac{t - t_0}{\tau}~.$$The reason this is frustrating is not that the hyperbolic tangent $\tanh$ is an awful function or anything -- it's actually very easy to work with and quite well-behaved! -- but because it means that we never actually attain the terminal velocity, we just get closer and closer to it. So the answer is trivially $\infty$.
With that said, if you want to find, say $v = 0.9 ~ v_t$ many calculators support the $\operatorname{atanh}$ operation that inverts the hyperbolic tangent; for example Wolfram Alpha gives $1.47222\dots$. With this time constant $\tau = \lambda / v_t$ you know that from rest, it takes $1.47~\tau$ to achieve that speed.
As for integrating once more to achieve a vertical position, it is not too hard to do that integral because $\tanh = \sinh / \cosh$ so you can simply use $u = \cosh~\frac{t - t_0}\tau$ for the substitution. So that gives a simple result of $$y(t) = y_0 - \lambda ~ \ln\left(\cosh~\frac{t - t_0}\tau \right).$$Hope that helps. Nonlinear differential equations rapidly get quite hairy; this is one of the few that is both really straightforward and which offers some really useful/invertible/manageable answers.