How to have a load pull more current from a low voltage high current battery
To start, well-insulated heated pants might not need 30-60 watts of power. I'd be concerned about overheating at that point, although I don't have a strong thermodynamics or bio/metabolism background to analyze this.

Assuming that the wire and battery are able to handle a larger current, you will want to either consider a shorter piece of wire, or multiple pieces in parallel. This will cause more current to be drawn from the battery, at roughly the same voltage (with a modest decrease due to load/output impedance). I'll continue with this assumption for the remainder of the answer, to keep math simple.

Given a battery voltage of 2.5 V (rounded for convenience), if you use 50 feet (5 ohms) of wire directly, you'll draw 0.5 amps for a total power of 1.25 watts (0.025 watt per foot).

If you use 25 feet of wire (2.5 ohms) you'll draw 1 amp, for total power of 2.5 watts (0.1 watt per foot). Now combine the two halves (25 feet each) in parallel, for a total of 2 amps and 5 watts.

Notice that if you split the wire into \$n\$ sections combined in parallel, you'll get \$n^2\$ times the current and thus \$n^2\$ times the power--thus, if you split the wire into six sections in parallel, you'll get roughly 45 watts, assuming that the battery and wire can handle the load: the current will be 18 amperes in total (3 amps per section). Each wire will need to handle 3 amps plus a safety factor, and will need to dissipate 7.5 watts over its length (almost a watt per foot). Make sure that the wires are suitably bonded to the load to effectively deliver heat, and that hotspots are avoided. 

Since these are heated pants, safety is paramount and hence you'll need to ensure that no burns occur.