In theory, if you had lossless conductors in the antenna and a lossless matching network, your shortened, 1 foot dipole would have a gain of only ~0.7 dB less on 160 meters than the gain of a full size 1/2 wavelength dipole. But the world is far from perfect.
The efficiency of an antenna is defined as:
$$\text {Efficiency}=\frac {R_r}{R_r+R_l} \tag 1$$
where $R_r$ is the radiation resistance of the antenna and $R_l$ is the resistive losses of the antenna.
As an antenna becomes shorter, its radiation resistance drops significantly. As you can see from formula 1, this causes the otherwise negligible resistive losses to significantly impact the efficiency of the antenna. Any inefficiency is simply dissipated as heat.
The gain of an antenna is defined as:
$$\text {Gain}=\text {Directivity}*\text {Efficiency} \tag 2$$
As the antenna is shortened, the directivity drops slightly from 1.64 for a 1/2 wavelength dipole to 1.50 for an infinitesimal dipole. This, combined with the lowered efficiency, results in an overall drop in gain.
So now let's plug your example into the above formulas.
The radiation resistance of a free space shortened dipole is given as:
$$R_r=197L_\lambda^2 \tag 3$$
where $L_\lambda$ is the length of the dipole expressed as a fraction of the wavelength (length in meters / 160 meters in this case).
Your 1 foot 160 meter dipole antenna example would therefore have an $R_r $ of only ~0.7 milliohms.
The RF resistance of 1 foot of 12 gauge wire at 1.8 MHz is ~17 milliohms (ref. http://chemandy.com/calculators/round-wire-ac-resistance-calculator.htm).
[Note: The following paragraph and the subsequent formula results are the result of Phil's critique of my original answer regarding resistive losses.]
The RMS current distribution in a very short dipole without end loading decays linearly from the maximum at the center point to zero at the ends of the antenna. As a result, the average RMS current of the short dipole is 1/2 that of the feedpoint current. Since resistive heat losses are the result of current squared, we must apply a 0.25 correction factor (0.5 squared) to the calculated calculated RF resistance of the wire in order to account for this RMS current slope along the length of the short dipole. The effective resistance of the wire in this shortened dipole is therefore ~4.3 milliohms.
Populating formula 1 tells us that the antenna will have a 14% efficiency. This means that 86 of your 100 watts of applied power goes to waste as heat. Formula 2 tells us that the antenna will have a free space gain of 0.21 or -6.8 dBi. For comparison, an efficient, full 1/2 wave dipole in free space has 2.2 dBi of gain.
Finally, a shortened antenna has a reactive component at its feedpoint that must be dealt with. This, in combination with a low resistive component, can be difficult to match to normal amateur transmission lines resulting in additional losses in the matching circuit and a corresponding reduction in antenna system gain. Of course if the matching circuit does not provide a match to the $Z_o$ of the transmission line, there will be additional losses in the transmission line and a possible reduction in the transmitter output power.
You mentioned dielectric losses. These are normally minimal at HF frequencies in a properly designed matching network.
