For this kind of problem, I find it easier to think about momentum than about mass and acceleration. Remember that momentum is mass times velocity,
$$
\vec p = m\vec v
$$
and so Newton's second law is secretly about the rate of change of momentum just as much as about acceleration:
$$
\vec F = m\vec a = \frac{\mathrm d}{\mathrm dt}(m\vec v)
= \frac{\mathrm d}{\mathrm dt} \vec p
$$
If your helicopter with mass $3000\,\rm kg$ were to suddenly fall out of the sky with acceleration $g = 10\rm\,m/s^2$, after each second it would gain momentum
$$
\begin{align}
\frac{\mathrm d}{\mathrm dt} \vec p &= m\vec g \\
&= 3000\,\mathrm{kg} \cdot 10\,\mathrm{m/s^2}
\\ &= 3\times10^4\,\frac{\rm kg}{\rm s}\,\frac{\rm m}{\rm s}
\end{align}
$$
I've done something a little sneaky with the units here: I've split up the $\rm s^{-2}$, to suggest that you can think of this rate of change of momentum as the product of a mass flow rate (usually $\dot m$, with a dot to mean it's a rate, in $\rm kg/s$) and an exhaust velocity ($\vec v_\text{out}$, in $\rm m/s$).
The mass flow rate is also proportional to the air flow rate: the faster the air is flowing, the more mass flow there is. If the density of the air is $\rho \approx 1\rm\,kg/m^3$, and the rotor blades have area $A \approx 100\rm\,m^2$ (which is just to keep the arithmetic simple, but winds up having roughly the same product $\rho A$ as your values) then the air flow rate would be
$$
\dot m = \rho A v
$$
If we want to find the rate of change of the momentum, we have to do a bit of calculus to get the flowing air from being at rest (above the rotor) to flowing at the exhaust speed:
$$
\frac{\mathrm d}{\mathrm dt} \vec p = \int_0^{v_\text{out}} \dot m \ \mathrm dv
= \rho A\cdot \frac12 v_\text{out}^2
$$
You might remember that the definition of pressure is force per unit area, $P = F/A$. So we can take this force $F = \mathrm dp/\mathrm dt$ and relate it to the pressure difference between the upper and lower edges of the helicopter's rotors:
$$
P_\text{below} - P_\text{above} = \frac{F_\text{net}}{A} = \frac12\rho \left(v_\text{below}^2 - v_\text{above}^2\right)
$$
This is actually just the Bernoulli equation,
$$
P + \rho gh + \frac12\rho v^2 = \text{constant}
$$
So you can say, with complete equivalence,
your helicopter hovers by transferring momentum to the air at the same rate that gravity would otherwise transfer momentum to the helicopter
your helicopter hovers by making the pressure below the rotor higher than the pressure above the rotor.
You can figure out the downdraft speed from these relationships, too:
\begin{align}
v_\text{out}^2 &= \frac{2 F_\text{net}}{\rho A}
\\
v_\text{out} &=
\sqrt{
\frac{2\cdot 3\times10^4\,\rm kg\,m\,s^{-2}}{1\rm\,kg\,m^{-3} \cdot 100\rm\,m^2}
}
\sim 25\rm\,m/s
\end{align}
This corresponds to a mass flow rate
$$
\dot m = \rho A v_\text{out} \approx 2500\rm\,kg/s
$$
just like you calculated.
I wouldn't think of this as a thirty-meter column of air above the rotor. Instead, I'd think of it as an order-of-magnitude estimate for how far below the helicopter you would expect to feel the downdraft. When the helicopter is on the ground (or hovering just above ground level), you can unquestionably feel the downdraft. As the helicopter rises, turbulence mixes the downdraft into a larger area $A$, so the average velocity of the downdraft decreases. If you have experience standing near helicopters as they take off or land, you might be able to estimate the shape of the cone where the downdraft is strong. If that cone were narrow, you might expect to feel the downdraft even when the helicopter is thirty meters up, as you say. But if that cone is wide, you would expect the turbulent downdraft from the helicopter to have dissipated by thirty meters below.
If you were to put a bigger rotor on your helicopter, giving a bigger flow area $A$, you would have larger mass flow rate, even at a smaller downdraft velocity.
