This is not a complete answer but it shows that in a "smaller" game, who will win can depend on the distribution of cards.
Take only the pile with four $2$s, four $3$s, four $4$s, four $4$s, four Aces. Arrange decks like this: First deck has four $2$s, second four $3$s, and so on. It is clear that the first player has winning strategy.
To see that the second can have a winning strategy, arrange decks like this:
- deck $A$ consist of three $2$s and an Ace
- deck $B$ consist of three $3$s and an Ace
- deck $C$ consist of three $4$s and an Ace
- deck $D$ consist of three $5$s and an Ace
- deck $X$ consist of $2$, $3$, $4$ and $5$
It will not be hard that the second player has winning strategy. I will write shortly what moves can happen ignoring symmetry.
Case I:. Player one takes $2$ from $X$, player two responds with Ace from $B$. Game over for player one.
Case II:. Player one takes Ace from $A$, player two responds with $3$ from $X$. Game over for player one.
So we are down to the only following case
Case III: Player one takes $2$ from $A$. Player two responds $3$ from $B$.
Case III.1. Player one takes $4$ from $C$, player two responds $5$ from $D$. Game over for player one.
Case III.2. Player one takes Ace from $C$, player two responds $5$ from $X$. Game over for player one.
Case III.3 Player one takes $4$ from $X$, player two responds Ace from $D$. Game over for player one.
This indicates that in the original game, probably both players can have winning strategies depending on the initial distribution of cards. Whether or not there is some "nice" property of distribution that will tell you who has the winning strategy, I don't know. I suspect the answer is "no" since most games, as verified by saulspatz end in draw.
Edit 1
It is easy to see that the original game can have both players as winners by expanding these distributions with decks of four $6$s, four $7$s, ..., four Kings. In the first arrangement it is clear that the first player can win. In the second whenever first player picks a card from first five decks, described as above, player two makes a counter move described as above. If first player chooses one of the remaining $8$ decks, player two chooses some other of the remaining $8$ decks, and since $8$ is even, player two will make the last move in those $8$ decks.
Edit 2
I wrongly imagined when it is a tie for the first part of the post, since in the first arrangement it is a tie, not win for player one. However, player one can have a winning strategy in $13$ cards game for the following configuration. Ace to $5$ as in example above. $6$ to $10$ same as Ace to $5$. And then four Jacks, four Queens, four Kings. Player one takes a Jack. At an point player two plays Ace to $5$, player one plays corresponding move in the same set. He will play the last move in that set as described above. Same for set $6$ to $10$. If player two takes Queen or King, player one takes the remaining of the two, so he is last in that set as well.
Considering this, it might not be possible to construct winning configuration for first player in smaller game, since he has to make two ranks unused in order to win.