I've put this as a second answer mainly for comparison with Johno's answer. I had a look at editing his answer, but it's quite long and laid out differently than I wanted to.

I have split the computation into four sections, based on the game stages:

1. Game setup - Fool's Landing is initially flooded and no one was dealt a special action card.
2. First Player's Actions - The first player is unable to shore up Fool's Landing.
3. First Player's Treasure Card Draw - Waters Rise is drawn (and Sandbags is not).
4. First Player's Flood Card Draw - Waters Rise is drawn a second time and sinks.

Notation:

• n The number of players
• v The number of Waters Rise cards drawn at the end of the first turn
• w The water level after the first turn

Game Setup

P(Fool's Landing drawn in first six cards) = 6/24
P(No sandbags or helicopters dealt) = (20 C (2n)) / (25 C (2n))

First Player's Actions

Let R be the first player's role:

P(R = Pilot and can't shore up Fool's Landing) = 1/6 * 0 = 0
P(R = Diver and can't shore up Fool's Landing) = 1/6 * 4611612/18574248 = 384301/9287124
P(R = Explorer and can't shore up Fool's Landing) = 1/6 * 10/69 = 5/207
P(R in {Navigator, Messenger, Engineer} and can't shore up Fool's Landing) = 3/6 * 59/138 = 59/276

So summing these:

P(Can't shore up Fool's Landing) = 1945439/6965343

First Player's Treasure Card Draw

For Fool's Landing to sink, one of the two drawn cards must be Waters Rise. The other must not be a sandbags. I'm keeping these separate as the value of v impacts the value of w.

P(v = 2 and no sandbags) = P(v = 2)
                         = 6 / ((28-2n)(27-2n))
P(v = 1 and no sandbags) = P(No sandbags | v = 1) * P(v = 1)
                         = (23-2n)/(25-2n) * 6(25-2n)/((28-2n)(27-2n))
                         = 6*(23-2n)/((28-2n)(27-2n))

First Player's Flood Card Draw

We have now shuffled the six flood cards and put them back on top of the Flood Card Deck. Note that w can be derived from the difficulty and the value of v.

P(Fool's Landing drawn at end of first player's turn) = w/6

The Result

The difficultly level impacts w:

• Novice: v=1 => w=2, v=2 => w=3

P(Fool's Landing Sinks First Turn) = 6/24 * (20C(2n))/(25C(2n)) * 1945439/6965343 * (6*(23-2n)/((28-2n)(27-2n)) * 2/6 + 6/((28-2n)(27-2n)) * 3/6)
 =  (20C(2n))/(25C(2n)) * 1945439/27861372 * (49-4n)/((28-2n)(27-2n))

• Normal/Elite: v=1 => w=3, v=2 => w=3

P(Fool's Landing Sinks First Turn) = 6/24 * (20C(2n))/(25C(2n)) * 1945439/6965343 * (6*(23-2n)/((28-2n)(27-2n)) * 3/6 + 6/((28-2n)(27-2n)) * 3/6)
 =  (20C(2n))/(25C(2n)) * 1945439/27861372 * (72-6n)/((28-2n)(27-2n))

• Legendary: v=1 => w=3, v=2 => w=4

P(Fool's Landing Sinks First Turn) = 6/24 * (20C(2n))/(25C(2n)) * 1945439/6965343 * (6*(23-2n)/((28-2n)(27-2n)) * 3/6 + 6/((28-2n)(27-2n)) * 4/6)
 =  (20C(2n))/(25C(2n)) * 1945439/27861372 * (73-6n)/((28-2n)(27-2n))

Finally, computing these formulae for different difficulty levels and number of players I ended up with:

Difficultly  | n |           Probability
Novice       | 2 | 1355970983/682632941760  ≈ 0.00199
Novice       | 3 | 1223681131/999834661980  ≈ 0.00122
Novice       | 4 |  429942019/608770978200  ≈ 0.00071
Normal/Elite | 2 |   33072463/11377215696   ≈ 0.00291
Normal/Elite | 3 |   99217389/55546370110   ≈ 0.00179
Normal/Elite | 4 |  859884038/837060095025  ≈ 0.00103
Legendary    | 2 | 2017420243/682632941760  ≈ 0.00296
Legendary    | 3 |   33072463/18178812036   ≈ 0.00182
Legendary    | 4 | 3009594133/2869920325800 ≈ 0.00105