The game is a draw. Specifically, with six-sided dice, the best winning probability that the first player can guarantee is $0.5$.
Assume that Aethelred and Brunhilde are playing the game with $n$-sided dice with Aethelred going first.
After the two players have finished picking their stickers, Aethelred has a length-$n$ subset $A$ of $\{1, \dotsc, 2n\}$ while Brunhilde has the complementary subset $B$.
A game then consists of chosing $i$ from $A$ and $j$ from $B$ and checking which is bigger.
Since all such pairs occur with equal probability $1/n^2$, the probability that Aethelred wins is
\begin{align*}\frac{\# \{(i, j) : i > j, i \in A, j \in B\}}{n^2}\end{align*}
(notice that draws are impossible, since $A$ and $B$ have no elements in common).
So for example, if $n=3$, one possible way in which the game plays out ends with $A = \{1, 4, 5\}, B = \{2, 3, 6\}$, in which case $\Pr(\text{Aethelred wins}) = 4/9$.
Using this idea, we can build a minimax tree for the game.
We encode a game state as a disjoint partition $(S_1, S_2, R)$ of $\{1, \dotsc, 2n\}$, where $S_1$ is the current set of labels of the player that is about to take their turn, $S_2$ the other player's set of labels and $R$ is the set of remaining labels.
A move then consists of the current player taking a subset $T$ of the remaining indices, adding them to their own set, and then switching the turn.
That is, the game enters the state $(S_2, S_1 \cup T, R \setminus T)$.
To score the game, we'll say that in a final position with no remaining labels
\begin{align*} \mathsf{score}(S_1, S_2, \{\}) &= \# \{(i, j) : i > j, i \in S_1, j \in S_2\} - \# \{(i, j) : i < j, i \in S_1, j \in S_2\} \\ &= n^2(2\Pr(\text{Current player wins from state } (S_1, S_2, \{\})) - 1) \end{align*}
so that the probability of the current player winning from state $(S_1, S_2, \{\})$ is
$$\frac{1}{2} + \frac{\mathsf{score}(S_1, S_2, \{\})}{2n^2}$$
and so that $\mathsf{score}(S_2, S_1, \{\}) = -\mathsf{score}(S_1, S_2, \{\})$.
Positions with a nonempty $R$ are scored under the assumption that the opponent always plays perfectly, maximising their position's score.
That is, after a valid move $T$, the current player's score is $-\mathsf{score}(S_2, S_1 \cup T, R\setminus T)$, so the current player maximises score according to
$$\mathsf{score}(S_1, S_2, R) = \max_{T \text{ is a valid move}} [-\mathsf{score}(S_2, S_1 \cup T, R\setminus T)].$$
This is now a standard minimax problem that can be solved using a minimax search tree.
I've implemented a (very badly optimised) version here, abailable on Try it Online.
Running the above code with $n = 6$, we find that the score of the initial position $(\{\}, \{\}, \{1, \dotsc, 12\})$ is $0$, and so the first player (Aethelred) has a probability of $0.5$ of winning assuming perfect play.
One possible perfect game consists of Aethelred and Brunhilde each selecting one sticker at a time, resulting in
\begin{align*}A &= \{5, 3, 12, 2, 10, 7\}\\B &= \{6, 4, 11, 8, 9, 1\},\end{align*}
where each label is picked in the order presented.
Notice that there are many possible perfect games.
