I'll first try to independently reproduce your calculation:
The thrust force is the transfer of momentum per unit time:
$$F = \frac{dp}{dt}.$$
Assuming ions accelerate by mutual Coulomb repulsion with the device without any interference, the magnitude of the momentum transferred to the device for each ion is equal to the ion's momentum, which we can get from the possible kinetic energy $T$ of 10 keV (10 kilovolts times a 1 electron charge).
$$v^2 = \frac{2T}{m} = \frac{2 \times 10^{4} \text{eV}}{28 \times 931 \times 10^{6} \frac{\text{MeV}}{c^2}} = {7.7 \times 10^{-7}}\ c^2$$
That makes $v$ equal to about $8.8 \times 10^{-4}\ c$, or $262,590\ \frac{\text{m}}{\text{s}}$, spot-on with your velocity number!
The thrust is equal to the mass and velocity of one nitrogen ion times the number of ions per second, for which you are using the maximum current of the transformer:
$$F = \frac{dp}{dt} = m_{N_2} \ v \ \frac{dn}{dt} = m_{N_2} \ v \ I \ \times \ 6.2 \times 10^{18},$$
where $~6.2 \times 10^{18}$ is 1 coulomb, the number of charges per second at the current of 1 ampere.
Switching back to MKS, 1 AMU is $1.66\times 10^{-27}\ \text{kg}$ and the current $I$ that you used in your calculation is $30\ \text{mA}$, so
$$F = 28 \times 1.66\times 10^{-27} \times 262,590 \times 30\times 10^{-3} \times 6.2 \times 10^{18},$$
and that gives 0.0023 newtons or about 100 times less than your value for the force of a sheet of paper of 0.178 newtons.
One problem with your solution is to assume all of the possible electrical power available from the transformer is used for ionization. There are many places for power to go here, so that's not a safe assumption, and your value of $5.59 \times 10^{-5} \frac{\text{kg}}{\text{s}}$ of ionized $\require{mhchem}\ce{N_2^+}$ is probably way too high. One $10\ \text{keV}$ electron is not likely to efficiently ionize hundreds of nitrogen molecules as it slowly decelerates. While cascade ionization does happen in a spark or breakdown, it's not close to 100% efficient, and in this situation there is no breakdown at all.
Instead, you can conserve charge or current, and treat the maximum possible thrust as coming from one ion produced for every electron collected from the nail, which is what I've done above.
That's if you are really accelerating 0.03 coulombs of nitrogen ions per second from $10\ \text{keV}$ potential all the way to ground. Much of the current may not be used, much of what little current you are using (probably tens or hundreds of micro-amperes if I had to guess) is going to coronal discharge and heating the air rather than producing thrust, but that's a different topic, as is my additional comments below about operating an ion thruster at ambient atmospheric pressure.
Comment on operation in atmosphere
For impulse or thrust calculations for nozzled engines or ion engines, the exhaust or reaction mass are moving at (roughly) full velocity when they leave the nozzle or exit grid so it's okay in those cases to use the velocity. But in your experiment, the ions never get a chance to accelerate to anywhere near their possible $10\ \text{keV}$ kinetic energy because you don't have an exit grid and a near-vacuum-filled acceleration space before it.
The velocity calculation resulting in $262,330 \frac{\text{m}}{\text{s}}$ assumes the ionized molecule accelerates without colliding with any other molecules along a path that goes all the way to ground potential (or nearly) and this would be several centimeters, but the mean free path in ambient air is only of the order of 0.1 microns. In this case thrust is the electrostatic repulsion between the nail and the (drifting) cloud of ions that the air has rapidly slowed to a drift velocity.
Below: An example of one kind of ion thruster. Acceleration takes place between the last two grids, before exiting the engine. In this case the final grid is even slightly negative. From here.
