There is an additional twist in the case where $\alpha$ is $\Sigma^0_1$. Assuming $\mathsf{WKL}_0$ (Weak König Lemma), $\forall Y\alpha(X,Y,n)$ is equivalent to a $\Sigma^0_1$ statement and hence so are $\exists n \forall Y \alpha(X,Y,n)$ and $\forall X\exists n \forall Y\alpha(X,Y,n)$.
The reason why $\exists X\phi(X)$ remains $\Sigma^0_1$ can be seen as follows. The verification that $\phi(X)$ holds for a specific $X$ can only use a finite amount of information about the set $X$. It follows that if we identify subsets of $\mathbb{N}$ with their characteristic functions, then $\{X \subseteq \mathbb{N} : \phi(X)\}$ is an open set in $2^{\mathbb{N}}$. Because $2^{\mathbb{N}}$ is compact, $\forall X\phi(X)$ holds if and only if there is an $n$ such that $\phi(X)$ holds for every $X \subseteq \{0,\ldots,n-1\}$ and $\phi(X)$ only uses information about membership in $X$ for numbers less than $n$ (and therefore $\phi(X')$ also holds whenever $X = X' \cap \{0,\ldots,n-1\}$). Since subsets of $\{0,\ldots,n-1\}$ are easily coded using numbers $\{0,\ldots,2^n-1\}$ this means that $\forall X \phi(X)$ is equivalent to a $\Sigma^0_1$ formula $\exists n\forall x < 2^n\widehat{\phi}(n,x)$, where $\widehat{\phi}(n,x)$ can be effectively computed from the original formula $\phi(X)$.
Note that this doesn't work if one uses functions $\mathbb{N}\to\mathbb{N}$ instead of subsets of $\mathbb{N}$ since Baire space $\mathbb{N}^{\mathbb{N}}$ is far from compact.