The answer is no.

Consider first the following version of the question. Regarding your Turing machine $T$ as a parameter, let $\varphi_{T,m}$ be the statement: $$\forall n_1\exists n_2\forall n_3\cdots\exists n_{2m}\ \ T\ \text{ accepts }\ \langle n_1,n_2,\ldots,n_{2m}\rangle.$$

The question is whether $\varphi_{T,m}$ is an assertion about $T$ and $m$ that exists in the Kleene hierarchy. In other words, does the set $S=\{(T,m)\mid \varphi_{T,m}\}$ have arithmetic complexity?

The answer to this version of the question is no. This set has hyperarithmetic complexity, which is on top of the Kleene arithmetic hierarchy. The reason is that it is essentially a truth predicate for arithmetic truth. For any $\Sigma_{2m}$ sentence $\forall n_1\cdots\exists n_{2m}A(\vec n)$, where $A$ is $\Delta_0$, let $T$ be a Turing machine accepting the satisfying instances of $A$, and thus $\theta$ is true exactly when $\varphi_{T,m}$. It follows that one can compute all the iterated jumps $0^{(n)}$ from an oracle for $S$, and thus it has higher complexity than $\Sigma_n$ for any particular $n$. Meanwhile, the set $S$ is computable from the omega jump $0^{(\omega)}=\oplus_n 0^{(n)}=\{\langle n,x\rangle\mid x\in 0^{(n)}\}$, since we can decide $\Sigma_n$ truth uniformly from $0^{(n)}$. Thus, the set $S$ is Turing equivalent to $0^{(\omega)}$. This oracle is Turing equivalent to the set $\{{}^\ulcorner\varphi^\urcorner \mid \mathbb{N}\models\varphi\}$, as $\varphi$ ranges over the arithmetic assertions. This set is known as TA, for ``true arithmetic''.

The succinct way to say it is: arithmetic truth is not arithmetic.

Now let's turn to your actual question, which is about the assertion $\exists m\ \varphi_{T,m}$, where you have quantified over $m$. Thus, your question is inquiring about the complexity of the set $\{T\mid \exists m\ \varphi_{T,m}\}$. The argument above gives the hyperarithmetic iterated jump $0^{(\omega+1)}$ as an upper bound on the complexity of this, since we can compute arithmetic truth from $0^{(\omega)}$, and you've added one more quantifier on top of this. Conversely, let's prove that the statement is not arithmetic. For any arithmetic sentence $\theta=\forall n_1\exists n_2\cdot \exists n_{2k}\psi(\vec n)$, where $\psi$ is $\Delta_0$, let $T$ be a Turing machine that accepts $\vec x$ just in case $\vec x$ has length $2k$ and $\psi(\vec x)$ holds. Thus, $\varphi_{T,m}$ holds if and only if $m=k$ and $\theta$ holds. Thus, your statement $\exists m\ \varphi_{T,m}$ holds if and only if $\theta$ holds. Thus, we have reduced TA to your set, and so your set is strictly in the hyperarithmetic hierarchy. So the answer is no.