Contrary to a comment I've now deleted, the representation is reducible. It's important to be clear about the action of $S_n$: given a representation $\rho: S_n \rightarrow \mathrm{GL}(V)$, composing with the quotient homomorphism $S_d \wr S_n \rightarrow S_n$ gives a representation of $S_d \wr S_n$, denoted $\mathrm{Inf}_{S_n}^{S_d \wr S_n} V$.

Using this, I'll show more generally that if $U$ is any representation of $S_d \wr S_{n-m}$ and $V$ is any representation of $S_d \wr S_m$ where $1 \le m < n$, then

$$ \begin{split} (U \boxtimes V)\bigl\uparrow_{S_d \wr S_{n-m} \times S_d \wr S_m}^{S_d \wr S_n} & \otimes \; \bigl( \mathrm{Inf}_{S_n}^{S_d \wr S_n} S^{(n-1,1)} \bigr) \\ &= \Bigl( (U \boxtimes V) \otimes \bigl( \mathrm{Inf}_{S_n}^{S_d \wr S_n} S^{(n-1,1)}\bigl\downarrow_{S_d \wr S_{n-m} \times S_d \wr S_m} \bigr)\Bigr)\bigl\uparrow^{S_n} \end{split}$$

is reducible. Since restriction commutes with inflation,

$$\bigl( \mathrm{Inf}_{S_n}^{S_d \wr S_n} S^{(n-1,1)} \bigr) \bigl\downarrow_{S_d\wr S_{n-m} \times S_d \wr S_m}
\, \cong \mathrm{Inf}^{S_d \wr S_{n-m} \times S_d \wr S_m}_{S_{n-m} \times S_m} \bigl( S^{(n-1,1)} \bigl\downarrow_{S_{n-m} \times S_{m}} \bigr). $$

Since $S^{(n-1,1)}\!\!\downarrow_{S_{n-m} \times S_m}\, \cong \bigl( F \boxtimes F \bigr) \oplus \bigl( F \boxtimes S^{(m-1,1)} \bigr) \oplus \bigl( S^{(n-m-1,1)} \boxtimes F\bigr)$ is reducible (this follows from Young's rule using Frobenius reciprocity, or the more general Littlewood–Richardson rule, or could be proved using an explicit basis for $S^{(n-1,1)}$), so is the right-hand side immediately above, and hence so is the original module. A similar argument will work replacing $S^{(n-1,1)}$ with $S^{(n-2,1,1)}$.