Consider the previous NSA periodical puzzle:
https://www.nsa.gov/newsfeatures/puzzlesactivities/puzzleperiodical/2016/puzzleperiodical04.shtml I haven’t looked at the official answer, nor anyone else’s.
Fun pages to start thinking about this problem are here, with a widget then generates a diagram for arbitrary sizes of rectangles and circles: http://www.engineeringtoolbox.com/circleswithinrectangled_1905.html and here http://www.datagenetics.com/blog/june32014/index.html.
My tentative answer is as follows:
 The description of the puzzle does not constrain all possible actions of the players (unlike the card puzzle discussed on another topic):
a. For example, it doesn’t say whether a player can move the existing coins around to make room for his coin when it is his turn to place a coin. If we assume that he isn’t allowed to do that, then I can’t find an answer, since there are too many ways of placing a coin. I therefore assume that he is allowed to move them about, while keeping them entirely on the table and not overlapping.
b. Similarly, I assume that he is not allowed to remove existing coins from the table, nor to stack them, because if that were possible then the game has no conclusion (it would go on forever) and so there is no best strategy.
 We assume an unlimited supply of quarters, more than could probably fit into a coffee can.
With these assumptions, we can proceed as follows.
I don’t have a proof of this strategy (yet, maybe ever). But I’m motivated by the following considerations.

We want to know if the maximum number of coins that could be packed on the table is odd or even. If it is odd then Dylan should elect to go first, and if it is even the he should go second.

The densest packing of small circles in a large rectangle (that can hold more than 30 coins) is achieved by hexagonal packing.

Hexagonally packed circles cover a plane with an efficiency of (pi * sqrt(3) / 6) ~ 90.69%

A US quarter is 0.955 inches in diameter, with an area = 0.7163028 in^2

Measure the length of the table, x, in inches, then its area is x^2.

The maximum number of quarters is x^2 * 90.69% / 0.7163028, rounded down to the nearest integer.
That last step makes a leap that I can’t really justify, that all the coins can be arranged hexagonally right up to the moment the last one is placed, but it seems right. Look at an example diagram from the widget I linked to (the lower square with the hexagonal packing, not the upper square with the square packing):
As long as you keep them in such offset rows, starting with one row along the top edge of the table, you will get there eventually.
The more dangerous assumption that I make is that the errors in my approximations won’t allow enough room for an extra coin to be squeezed in.