Do I look for a low signal relay? like .03VAC to 12VDC?
You certainly can
look for one, but without knowing how much voltage (presumably AC since the speaker must make a "ringing" sound) and how much current is available from the cell phone to operate it, what are you going to look for?
Back in the day (I am talking the previous century, and very early in that), one could obtain sensitive DC coil relays that operated at low coil voltage and small coil currents. By low coil voltage, I mean less than ten volts. By small coil currents, I mean one mA (1000 μA) or less. You do need to rectify, using four diodes in a full-wave bridge configuration, the AC "ringing" signal from the telephone before applying the rectified (now DC) voltage to the coil of a suitable relay. Nothing further, other than a circuit for your "loud ass horn" in series with the normally-open relay contacts and a power supply for the horn, is required.
So, wire up four 1N400x-series diodes as a full-wave bridge (FWB) rectifier, or purchase a ready-made FWB module. Connect the AC output of the cell phone to the bridge AC input, and connect the relay coil to the bridge output. Turn up the volume on the phone until the relay contacts pull in when the phone rings. Disconnect the 8Ω speaker for more output voltage if necessary or desirable to mute the audible ringing. The 1N400x diodes can be anything in that series: 1N4001 ... 1N4007 will all work. Buy the cheapest ones you can find. The higher part numbers represent diodes with greater peak inverse voltage (PIV) ratings, but even the lowest PIV rated diodes will work just fine. The key component is the relay and its requirements for coil voltage and coil current required to actuate the relay contacts.
All this means you should build (or buy) the bridge rectifier FIRST, connect it to the two ringer wires from the phone, and then measure how much DC output voltage and DC output current you can obtain with the phone ringing and the volume turned up to maximum. Put a temporary resistive load, somewhere between 100Ω and 1000Ω, across the diode bridge output and make DC voltage measurements. Use I = E / R to measure how much current, I, is available to operate the relay coil at whatever voltage, E, and resistance, R, you have as a load. Remember when using Ohm's Law that I is measured in amperes, E in volts, and R in ohms. Therefore, if you measure ten volts across a thousand ohm resistor, there is ten milliamperes (0.010 A) of current in the resistor.
Of course, with just a little bit of effort, you can use just about
any relay coil (AC or DC!) by adding some signal processing components, but that hardly seems necessary if you can find the right relay for the job. The KISS principal applies here. So make the voltage and current measurements (under load!) first, then go looking for a relay whose maximum pull-in voltage and whose minimum pull-in current is less than your measured values. Of course, it is highly unlikely that the cell phone will deliver ten volts across one hundred ohms (0.1 A current in the resistor), or need less than one volt across a thousand ohms (0.001 A current in the resistor) to actuate the contacts of a very sensitive relay.
When making the voltage measurements (current is calculated, not measured, based on the resistance and the voltage measured across it), it may be necessary to filter the DC output voltage from the bridge rectifier to obtain stable readings. To do this, place a 100μF (or greater valued) polarized electrolytic capacitor across the DC bridge output, in parallel with the load resistor. Make sure the negative lead of the electrolytic capacitor is connected to the negative output terminal of the bridge rectifier. It does no harm to leave this capacitor permanently connected across the bridge rectifier output, but it will introduce a slight delay between the phone ringing and the horn sounding as the capacitor charges while the phone is ringing.
Please let us know if you get your phone to blow your horn!