OK, there's a lot of steps involved in the calculation, but it's really pretty straightforward. My apologies for not doing figures here, but I'm not on my home computer. I'll normalize things to the cartridge output and then put units on at the end. So for simplicity, the cartridge output is 1. That transformer has two turns ratios, 1:8 and 1:16, the former achieved by putting the primaries in series, the latter by putting the primaries in parallel. Recall that for a turns ratio of N, the voltage transforms as N, the impedances transform as N^2.
As a rough guide, at standard velocity levels (3.54 cm/s, 1 kHz), the cartridge outputs 0.3mV, so for an ideal 1:8, the output would be 2.4mV. For an ideal 1:16, 4.8mV. The actual levels will be slightly below this (10% below is a rule of thumb), but neither are likely to cause overload or insufficient signal problems. So unless you're anal, stop here. Say 4.3mV or so for the 1:16 stepup.
Let's look at it a bit more exactly. First, the primary side. We'll use 1:16 as our example. The primaries in parallel are 1.8 ohm/2 = 0.9 ohm DCR. The secondaries in series have a DCR of 105x2 = 210 ohm. The secondary impedance is reflected back to the primary as (DCR of the secondary x 2 since they're in series + load resistance which is probably 47k)/N^2. Putting in numbers, the reflected secondary impedance is then (210+47000)/16^2 = 184 ohm. So that's the load the cartridge sees.
Now, the DCR of the cartridge forms a voltage divider with the primary impedance. Since the primary DCR is under an ohm, it's negligible compared to that 184 ohm, so we'll neglect it. The voltage delivered to the transformer for the assumed unity output then is effectively 184/(10 + 184) = 0.95. So we've lost a little bit.
On the secondary side, the reflected primary impedance is (10 + 0.9)N^2 = 2790 ohms. (The numbers represent the cartridge DCR plus the primary DCR). This, in series with the secondary DCR of 210 ohm, form a voltage divider with the preamp input of 47k. So... we have 0.95 (calculated above) times 16 (the turns ratio) in voltage driving that divider. Effectively the preamp sees (0.95)(16)(47000)/(47000+2790) = 14.35 for that input of 1.
So... for your cartridge output of 0.3mV, the voltage the preamp would actually see is (14.35)(0.3mV) = 4.305mV, which is what the simple rule-of-thumb predicted.