Electronic – Higher than expected leakage drain current with Vgs=0 for N channel MOSFET

The use of a MOSFET for reverse voltage protection is very straight forward.
Some of your references are correct but of low relevance and are tending to make the problem look more complex than it is. The key requirements (which you have essentially already identified) are

• MOSFET must have enough Vds_max rating for maximum voltage applied

• MOSFET Ids_max rating more than ample

• Rdson as low as sensibly possible.

• Vgs_max not exceeded in final circuit.

• Power dissipation as installed able to sensibly handle operating power of I_operating^2 x Rdson_actual

• Power dissipation as installed able to handle turn on and off higher dissipation regions.

• Gate driven to cutoff "rapidly enough" in real world circuit.
  (Worst case - apply Vin correctly and then reverse Vin instantaneously. Is cutoff quick enough?)

In practice this is easily achieved in most cases.
Vin has little effect on operating dissipation.
Rdson needs to be rated for worst case liable to be experienced in practice. About 2 x headlined Rdson is usually safe OR examine data sheets carefully. Use worst case ratings - DO NOT use typical ratings.

Turn on may be slow if desired but note that dissipation needs to be allowed for.
Turn off under reverse polarity must be rapid to allow for sudden application of protection.

What is Iin max ?
You don't say what I_in_max is and this makes quite a difference in practice.

You cited:

"If the drain-to-source voltage is zero, the drain current also becomes zero regardless of gate–to-source voltage. This region is at the left side of the VGS– VGS(th)= VDS boundary line (VGS – VGS(th) > VDS > 0).

and

Even if the drain current is very large, in this region the power dissipation is maintained by minimizing VDS(on)."

Note that these are relatively independent thoughts by the writer. The first is essentially irrelevant to this application.
The second simply says that a low Rdson FET is a good idea.

You said:

Does this configuration fall under the VDS = 0 classification? That seems like a somewhat dangerous assumption to make in a noisy environment (this will be operating in the vicinity of various types of motors), as any voltage offsets between input supply ground and local ground could cause current to flow. Even with that possibility, I'm not sure I need to spec for my maximum load current on the drain current ID. It would then follow that I don't need to dissipate very much power either. I suppose I could mitigate the problem by Zener clamping VGS closer to VGS(th) to reduce drain current/voltage?

Too much thinking :-).

When Vin is OK get FET turned on asap.
Now Vds is as low as it is going to get and is set by Ids^2 x Rdson
Ids = your circuit current.
At 25C ambient Rds will start at value cited at 25C in spec sheet and will rise if/as FET heats. In most cases FET will not heat vastly.
eg 1 20 milliOhm FET at 1 amp gives 20 mW heating. Temperature rise is very low in any sensible pkg with minimal heatsinking. At 10A the dissipation = 10^2 x 0.020 = 2 Watts. This will need a DPAk or TO220 or SOT89 or better pkg and sensible heatsinking. Die temperature may be in 50-100C range and Rdson will increase over nominal 25C value. Worst case you may get say 40 milliOhm and 4 Watts. That is still easy enough to design for.

Added: Using the 6A max you subsequently provided.
PFet = I^2.R. R = P/i^2.
For 1 Watt disspation max you want Rdson = P/i^2 = 1/36 ~= 25 milliohm.
Very easily achieved.
At 10 milliohm P = I^2.R = 36 x 0.01 = 0.36W.
At 360 mW a TO220 will be warm but not hot with no heatsink but good airflow. A trace of flag heatsink will keep it happy.

The following are all under $1.40/1 & in stock at Digikey.

LFPACK 60V 90A 6.4 milliohm !!!!!!!!!!!

TO252 70V 90A 8 milliohm

TO220 60V 50A 8.1 milliohm

You said:

I suppose I could mitigate the problem by Zener clamping VGS closer to VGS(th) to reduce drain current/voltage?

No!
Best saved for last :-).
This is the exact opposite of what is required.
Your protector needs to have minimal impact on the controlled circuit.
The above has mjaximum impact and increases dissipation in protector over what can be achieved by using a sensibly low Rdson FET and turning it on hard.

From the data sheet Fig. 2, you can see a curve at different VGS values. Using the lowest curve (3.0 V), you can see it crosses 2 A at 0.2 V (it's a log-log scale). Therefore the on-resistance is 0.2/2 = 0.1 ohm. This is at +150 C, which is also the maximum allowed junction temperature (this is quite hot and would burn your finger if you touched it). Note that this is actually a typical plot, not a guaranteed worst case value, but in practice FETs are within 20 % of this value. Therefore let's assume 0.12 ohm.

If you don't have a real heatsink, it might be reasonable to assume a thermal resistance of 100 deg/W (with it mounted on a large sheet of metal, it might be 30 deg/W or less). Assuming an ambient of 60 deg (also very warm), and allowing up to 150 C junction, you can have a temperature rise of 150-60 = 90 deg. At 100 deg/W, this allows 90/100 = 0.9 W dissipation.

Using P = I^2*R, and 0.12 ohm gives a current of sqrt(0.9/0.12), or just over 2 A.

The 12 V LED strip value doesn't directly matter (as long as the FET's max VDS rating exceeds this which it does).

In cases like this, it's always better to be on the safe side and round calculations to the 'safe' side.