Very small transformers are really lousy and primary resistance can be a lot.
For any transformer over about 50 VA the efficiency will be 80% or better at full load. That means that at full load the losses are less than 10 W at full load for a 50 VA transformer.
The primary resistive loss is I^2 * R so on 120 V ac input, the primary current is about 0.42 amps, so R can't be more than 57 Ω, and that would be with all the losses in the primary. It is more likely to be less than about 30 Ω.
You have done this calculation based on the assumption that the transformer is under full load, and you haven't accounted for the fact that a transformer that can supply 50 VA to a load will absorb more than 50 VA at the primary under load. The primary current under full load will be more like .5 amps.
The primary current with no load will be about 0.1 A, so the impedance is around 1200 j Ω. The "j" indicates reactive impedance, effectively at right angles to the resistive impedance. The 30 Ω would only increase it to sqrt(1200^2 + 30^2) = 1200.37 Ω
But now you are considering the unloaded transformer, so the 30 Ω is no longer the relevant resistance.
What that means is that the primary resistance makes no practical difference to the primary current at no load. Any effect is lost in measurement errors.
Here are some measurements on a 50 VA transformer (25.2 VAC output at 2 amps).
Primary DCR = 9.5 Ω
Secondary DCR = .74 Ω
Measured turns ratio = 4.08
Rated secondary current = 2 amps
Rated primary current = .49 amps
With 120 VAC applied to the primary and the secondary open:
Primary current = .109 amps
Watts loss = 3.6 watts (measured with precision wattmeter)
Since the current is so low (compared to full load primary current), this loss figure is essentially all core loss, and maybe .1 watt copper loss.
We can calculate the unloaded primary impedance as 120/.109 = 1101 Ω
Let's resolve this into reactance and resistance components. First calculate the resistive component. Knowing that the loss is 3.6 watts, we have I^2*Reff = 3.6, so Reff = 303 Ω. Then using the relationship Z^2 = X^2 + R^2, we have X (the reactance) = SQRT(1101^2 - 303^2) = 1058.4 Ω.
From this we can calculate the effective unloaded primary inductance. Leff = 1058.4/(2*PI*60) = 2.81 henries.
To measure the copper loss, we short the secondary and with a variac apply gradually increasing voltage to the primary until the primary current is equal to the full rated primary current (.49 amps). In the case of the transfomer under consideration, this happens with about 10 VAC applied. With such a low applied primary voltage, the core loss is entirely negligible, and the measured loss is the copper loss under full load. We measure the loss under this condition with a precision wattmeter; the value is 5.5 watts. Strictly, the transformer should have been allowed to reach normal full load operating temperature to account for the increase in the copper resistance with elevated temperature, but I don't feel like waiting that long.
So, we have the core loss as 3.6 watts, and the copper loss as 5.5 watts, for a total transformer loss under load of 9.1 watts.
This means that the transformer loss resistance under full rated load is 9.1/(.49^2) = 37.9 Ω, compared to 303 Ω with the secondary unloaded.
In either case, the loss resistance is a small part of the overall primary impedance. Under full (resistive) load, though, the primary impedance is almost purely resistive (about 240 Ω) compared to the almost purely reactive primary impedance when the secondary is unloaded.