Let $A$ be a nonempty set and $X \subseteq A^{\mathbb{N}}$. In the classic version game runs in such a way: there are two players (I and II), at the beginning I writes any $a_1 \in A$, then II writes any $a_2 \in A$, I writes any $a_3 \in A$ etc. Eventually players get an infinite sequence $(a_n)$. I wins if $(a_n) \in X$, II wins in other case.
If any player has a winning strategy in the game, $X$ is called determined. According to Borel Determinacy Theorem, for any nonempty $A$ any Borel $X$ is determined.
Let's now describe a non-classic version of the game: $A=\{0, 1\}$ and players make any finite number of sequences simultaneously. For example, they make three sequences $(a_n), (b_n)$ and $(c_n)$, so I plays $a_1, b_1, c_1 \in \{0, 1\}$, then II plays $a_2, b_2, c_2 \in \{0, 1\}$, then I plays $a_3, b_3, c_3 \in \{0, 1\}$ etc. Now $X \subseteq {(\{0, 1\}^\mathbb{N})}^3$. I wins if $X$ contains all the three sequences, in other case II wins.
I want to understand the following thing: is Borel Determinacy Theorem still true for this non-classic version of the game?
Yes. In your case $A$ is just the set of binary sequences of length $3$ ($A = \{0,1\}^{\{0,1,2\}}$) and the underlying space is $A^{\mathbb{N}}$. Player I plays $3$ bits, then Player II and so on. Whenever $X \subseteq (\{0,1\}^{\mathbb{N}})^3$ is Borel, the corresponding subset of $A^{\mathbb{N}}$ is Borel as well, and vice-versa.