I think $[X] \frown \mathfrak{P}(m \oplus n) = 2\langle m \cdot n + n^3, [Y] \rangle$.

The class $m \oplus n$ is better thought of as $m \otimes 1 + n \otimes x$ under the Kunneth decomposition, where $x \in H^1(S^1;\mathbb{Z}/2)$ is the nontrivial element. Then the quadratic property of $\mathfrak{P}$ and naturality gives
$$\mathfrak{P}(m \otimes 1 + n \otimes x) = \mathfrak{P}(m) \otimes 1 + \mathfrak{P}(n \otimes x) + 2(m \smile n \otimes x).$$

Firstly, $\mathfrak{P}(m) = 0$ as $Y$ is 3-dimensional. 

Secondly, the fact that the suspension of the Pontrjagin square is the Postnikov square (and that the Postnikov square is not universally trivial, which bizarrely I can't find a reference for) means that
$$\mathfrak{P}(n \otimes x) = \bar{\mathfrak{P}}(n) = 2 n^3.$$

Remark: In an earlier version of this answer I had consulted

    Nakaoka, Minoru
    Note on cohomological operations. J. Inst. Polytech.
    Osaka City Univ. Ser. A. Math. 4, (1953). 51–58.

which has the formula
$$\mathfrak{P}(n \otimes x) = \mathfrak{P}(n) \otimes \mathfrak{P}(x) + \bar{\mathfrak{P}}(n) \otimes \beta(Sq_2(x)) + \beta(Sq_2(n)) \otimes \bar{\mathfrak{P}}(n)$$
where $\bar{\mathfrak{P}}(-)$ is the Postnikov square (i.e. the operation given on cochains by $u \mapsto u \cup \delta u$), and $\beta$ is the Bockstein to $\mathbb{Z}/4$-cohomology. Each of $\mathfrak{P}(x)$ and $\bar{\mathfrak{P}}(x)$ must be trivial by degree reasons. If one interprets $Sq_2(x)$ literally it also ought to be zero, but this is apparently wrong and it ought to be interpreted as $1$.