Is Scrabble's method of determining turn order fair?

Because of symmetry, no player has an advantage. That is, if $\mathscr A$ is the set of outcomes where player $A$ gets the letter closest to A and $\mathscr B$ is the set of outcomes where player $B$ gets the letter closest to A, there is a one-to-one correspondence between $\mathscr A$ and $\mathscr B$ obtained by interchanging the letters players A and B get. Since all outcomes are equally likely, they have the same probability of getting to go first.

Your intuition would make sense if you could choose individual tiles with unequal probabilities. For example, imagine that there are just two tiles, 'A' and 'B', and the selection process is such that tile 'A' is chosen with probability $0.75$ and 'B' with $0.25$. Then obviously the player who draws first has a higher probability of being the winner.

Now imagine that instead of two we have four tiles, and all are equiprobable to pick (much like your scrabble scenario). Three 'A's and one 'B'. Now you still have a $0.75$ probability of drawing 'A', but unlike the first case, you do not pick all the 'A's from the bag if you pick an 'A'.

What's the probability of winning then? Let's name the outcome X and give it a value $+1$ for winning and $-1$ for losing. What is the expected value $E(X)$ if you go first? If you are unlucky to draw the 'B' you certainly lost (there are only 'A's for the second person to pick). If you pick 'A', you might win (with probability 1/3 the second person will pick 'B'), or you might replay the whole game (because the second person will pick another 'A' and it's a draw). Let's write these down:

$$E(X) = 0.25\cdot (-1) + 0.75\cdot(\frac{1}{3}\cdot 1 + \frac{2}{3}\cdot E(X)) \iff \\ E(X) = -\frac{1}{4} +\frac{1}{4} + \frac{1}{2}E(X) \iff \\ E(x) = \frac{1}{2}E(X) \iff\\ E(X) = 0$$

You see how the good cases are balanced out by the bad ones and the net result is zero gain. Granted, this is a specific example to illustrate a point, but you can easily see how this generalises to any probabilities.

There is of course the symmetry argument outlined in the answer by Robert Israel, that generalises this for any number of players. This works because tiles are equiprobable, and I think this approach here helps us a little to get a better intuition.

How do the players draw their tiles? If the tiles are in a bag, and the players reach into the bag and pull out a tile, they may cheat by feeling for a blank tile. Or, if the tiles are spread out face down on the table, players may be able to recognize them by marks on the back. If this kind of cheating is allowed, the player who draws first has an advantage. However, IF WE ASSUME RANDOM DRAWING, no player has an advantage.

Suppose that the first player has a better chance of drawing a blank than the second player. Then, by the same token, the first player has a better chance of drawing a Z than the second player, because (under our assumption of random drawing) there is no difference between a blank and a Z (except that there are two blanks and only one Z in the standard English set). And the same goes for every other letter of the alphabet: the first player has a better chance of drawing it than the second player. But this is plainly absurd. It would mean that, over a long series of games, the first player would draw every tile in the set more often than the second player.