First, compute the power you need.
This will consist of radiated and conducted heat loss.
The radiated power can be computed from Stefan's Law. Assume an emissivity of 1.0 to be on the conservative side; you will also need the wire diameter in order to compute the radiative area.
This is likely to be relatively low. If you can keep the wire in a high vacuum, this will be the total power you need from the battery.
Conducted heat loss requires knowledge of the material surrounding the wire, such as air, water, insulating material, etc. Assuming still air, use Newton's Law. You need to plug in a "heat transfer coefficient" ... assuming the wire is horizontal, here's how to calculate it. For the value of k the thermal conductivity of air, use
k = 0.024 W/m/K. You will need the wire diameter D, you can assume a Prandtl Number of 0.8 and a Rayleigh number of about 1E6 as a starting point, and you must make reasonable assumptions about the air temperature.
If the air is moving, cooling will increase and things get much more complex and less accurately calculable as calculating the heat transfer coefficient gets more difficult.
If this computation is too much, bypass it by experimenting with a metre of wire and a variable power supply.
This is likely to give a much higher power requirement than radiative cooling at the temperature given.
Add these figures and you now have a power requirement. Given 12V, you can calculate the required current, and thus the life of a car battery.
You can also calculate the resistance you need as a load. Compare that with the resistance for 1km of your chosen diameter of Nichrome wire and you will see that you cannot power a continuous 1km length from only 12V. Which means you need multiple short lengths of wire connected in parallel, or a much higher voltage than 12V.
EDIT : The new information changes things A LOT. You'll achieve this temperature with much less power because the wires are in close proximity so they are heating each other. Model the heated surface as a plane when calculating radiation losses and heat transfer coefficients.
Parallel 2m wires are the way to go, and in a 3m^2 (2m*1.5m) blanket they will only be 3mm apart. Why parallel? Because with one single wire, a single break will stop the entire blanket, and be difficult to find and repair, while with parallel wires, a few breaks won't even be noticed. As well as allowing much safer voltages around your children!
You'll also have to revise that temperature downwards - if the wire's at 110C the surface of the blanket could be around 100C, causing burns.