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You might hope to be able to construct an arithmetic degree containing an aT-complete set above any arithmetic degree $X$. However, I think this is impossible. By the argument given $X$ there is an arithmetic degree $Z$ with $X <_a Z <_a X^{\omega}$ and $X <_T Z <_t X^{\omega +1}$ not containing an aT-complete set. Now consider the property $P(X): X$ there is an aT-complete set in the arithmetic degree of $X$. $P(X)$ is a Borel property that's invariant under Turing equivalence. Hence, by Martin's cone theorem the class of sets satisfying $P$ either contains or is disjoint from a cone. But since there is a $Z$ above any $X$ not satisfying $P$ there must be a cone in the Turing degrees whose members all fail to satisfy $P$.

I hypothesize that $0^{\omega}$ is the base of such a cone in the arithmetic degrees (we can turn any cone in the Turing degrees into a cone in the arithmetic degrees by considering representatives).

Note that the above argument can actually build an $A$ computable in $0^{\omega + n +1}$ witnessing that $0^{\omega +n}$ isn't aT-complete. Hence the arithmetic degree containing $0^{\omega}$ doesn't contain an aT-complete set.

You might hope to be able to construct an arithmetic degree containing an aT-complete set above any arithmetic degree $X$. However, I think this is impossible. By relativising the argument given above to $X$ we can show that the arithmetic degree of $X^{\omega}$ doesn't contain an aT-complete set. Hence, no upper cone (in Turing or arithmetic degrees) contains only sets whose arithmetic degree is an aT-complete set. Thus, by Martin's cone theorem (since this is a Borel, degree invariant property) there must be an upper cone none of whose members is arithmetically equivalent to an aT-complete set.

Actually, we can show that $0^{\omega}$ is the base of such a cone. Note that for any $C$ there is an $A$ with $A^{\omega} \equiv_T 0^{\omega} \oplus C$ (code $C$ into an $\omega$-generic). By the argument above the arithmetic degree of $A^{\omega}$ doesn't contain an aT-complete set.

Ok, I'm pretty sure the answer is no to the basic question (still not sure about the generalization). We build a set $A$ to satisfy the following conditions.

$P_e: A \neq \phi_i(0^{n})$ $R_{i,j}: X = \phi_i(0^\omega) \land (\forall i)\left(X(i) = A \models \psi_j(i)\right) \implies (\exists n)(X \leq_T 0^{n})$

Where $\phi_i$ is a Turing reduction and $\psi_j$ is an arithmetic formula with one free variable.

We build $A$ via sequence of perfect trees $T_n$ with $T_{n+1}$ a subtree of $T_n$ as in the construction of a minimal arithmetic degree to meet the above requirements with $\lvert T_n(0) \rvert \geq n$. We ensure that each $T_n$ is arithmetic and is uniformly computable from $0^{\omega + 1}$ and set $T_{-1} = 2^{< \omega}$

To meet $P_e$ we simply insist that $T_e(0)$ disagrees with $\phi_i(0^{n})$. Given an index for $T_{e - 1}$ we can find an index for a subtree $\hat{T}_e$ with this property and with $\lvert T_n(0) \rvert \geq e$ computably in $0^{\omega}$.

To meet $R_{i,j}$ with $e = \langle i,j \rangle$ now ask if there is a string $\tau$ on $\hat{T}_e$ and an $x$ such that $\tau \Vdash \psi_j(x) = k \land \phi_i(0^{n};x) = 1 -x$ (here we are using local forcing on $\hat{T}_e$). If so we let $T_e$ be the subtree of $\hat{T}_e$ above $\tau$ and otherwise let it be $\hat{T}_e$. Note that this whole construction can be carried out computably in $0^{\omega +1}$

It's obvious that the requirements $P_e$ are satisfied and we've constructed $A \leq_T 0^{\omega+1}$ so $A$ clearly satisfies $0 <_a A \leq_A 0^{\omega}$. If $R_{i,j}$ is also satisfied then we are done since any $\widehat{A} \leq_a A$ with $\widehat{A} \leq_T 0^{\omega}$ would necessarily be arithmetic and therefore $A$ couldn't be arithmetic in $\widehat{A}$.

So suppose that for some $i,j,n, X$ we have a $\Sigma^0_n$ formula $\psi_j$ defining $X$ from $A$ and $X = \phi_i(0^{\omega})$. Now since $\hat{T}_{\langle i,j \rangle}$ is arithmetic we can find an arithmetic path $Z$ through $\hat{T}_{\langle i,j \rangle}$ that's $n$-generic under local forcing. Thus, the set defined by $\psi_j$ applied to $Z$ is also arithmetic and by construction must be equal to $\phi_i(0^{\omega}) = X$ (or we'd have found a disagreement). Hence, $X$ is arithmetic (i.e. computable from $0^{m}$ for some $m$).

Ok, I'm pretty sure the answer is no to the basic question. We build a set $A$ to satisfy the following conditions.

$P_e: A \neq \phi_i(0^{n})$ $R_{i,j}: X = \phi_i(0^\omega) \land (\forall i)\left(X(i) = A \models \psi_j(i)\right) \implies (\exists n)(X \leq_T 0^{n})$

Where $\phi_i$ is a Turing reduction and $\psi_j$ is an arithmetic formula with one free variable.

We build $A$ via sequence of perfect trees $T_n$ with $T_{n+1}$ a subtree of $T_n$ as in the construction of a minimal arithmetic degree to meet the above requirements with $\lvert T_n(0) \rvert \geq n$. We ensure that each $T_n$ is arithmetic and is uniformly computable from $0^{\omega + 1}$ and set $T_{-1} = 2^{< \omega}$

To meet $P_e$ we simply insist that $T_e(0)$ disagrees with $\phi_i(0^{n})$. Given an index for $T_{e - 1}$ we can find an index for a subtree $\hat{T}_e$ with this property and with $\lvert T_n(0) \rvert \geq e$ computably in $0^{\omega}$.

To meet $R_{i,j}$ with $e = \langle i,j \rangle$ now ask if there is a string $\tau$ on $\hat{T}_e$ and an $x$ such that $\tau \Vdash \psi_j(x) = k \land \phi_i(0^{n};x) = 1 -x$ (here we are using local forcing on $\hat{T}_e$). If so we let $T_e$ be the subtree of $\hat{T}_e$ above $\tau$ and otherwise let it be $\hat{T}_e$. Note that this whole construction can be carried out computably in $0^{\omega +1}$

It's obvious that the requirements $P_e$ are satisfied and we've constructed $A \leq_T 0^{\omega+1}$ so $A$ clearly satisfies $0 <_a A \leq_A 0^{\omega}$. If $R_{i,j}$ is also satisfied then we are done since any $\widehat{A} \leq_a A$ with $\widehat{A} \leq_T 0^{\omega}$ would necessarily be arithmetic and therefore $A$ couldn't be arithmetic in $\widehat{A}$.

So suppose that for some $i,j,n, X$ we have a $\Sigma^0_n$ formula $\psi_j$ defining $X$ from $A$ and $X = \phi_i(0^{\omega})$. Now since $\hat{T}_{\langle i,j \rangle}$ is arithmetic we can find an arithmetic path $Z$ through $\hat{T}_{\langle i,j \rangle}$ that's $n$-generic under local forcing. Thus, the set defined by $\psi_j$ applied to $Z$ is also arithmetic and by construction must be equal to $\phi_i(0^{\omega}) = X$ (or we'd have found a disagreement). Hence, $X$ is arithmetic (i.e. computable from $0^{m}$ for some $m$).

However, there are 'non-trivial' arithmetic degrees containing aT-complete sets if by non-trivial we mean having infinitely many predecessors. Consider the usual construction of a Turing degree $\mathbf{d}$ with $\mathscr{D}(<, \mathbf{d})$ equal to $\omega$. We build $A$ using uniform trees $T_n$ where if $\lvert \sigma \rvert = \langle m, i \rangle$ then $T_n(\sigma\hat{}0)$ and $T_n(\sigma\hat{}1)$ only disagree on locations of the form $\langle i, x \rangle$. We then perform the construction so that $A_0 = A^{[0]}$ is a minimal degree, $A_1 = A^{[\leq 1]}$ is a strong minimal cover of $A^{[0]}$ and so on with any set computable from $A$ either computable in some $A_i$ or computing $A$.

Now modify this construction to use local forcing and arithmetic rather than computable trees as in the construction of a minimal arithmetic degree. As $A$ explicitly computes a representative $A_i = A^{[\leq i]}$ of every arithmetic degree below $A$ it's aT-complete.

You might hope to be able to construct an arithmetic degree containing an aT-complete set above any arithmetic degree $X$. However, I think this is impossible. By the argument given $X$ there is an arithmetic degree $Z$ with $X <_a Z <_a X^{\omega}$ and $X <_T Z <_t X^{\omega +1}$ not containing an aT-complete set. Now consider the property $P(X): X$ there is an aT-complete set in the arithmetic degree of $X$. $P(X)$ is a Borel property that's invariant under Turing equivalence. Hence, by Martin's cone theorem the class of sets satisfying $P$ either contains or is disjoint from a cone. But since there is a $Z$ above any $X$ not satisfying $P$ there must be a cone in the Turing degrees whose members all fail to satisfy $P$.

I hypothesize that $0^{\omega}$ is the base of such a cone in the arithmetic degrees (we can turn any cone in the Turing degrees into a cone in the arithmetic degrees by considering representatives).

Ok, I'm pretty sure the answer is no. We build a set $A$ to satisfy the following conditions.

$P_e: A \neq \phi_i(0^{n})$ $R_{i,j}: X = \phi_i(0^\omega) \land (\forall i)\left(X(i) = A \models \psi_j(i)\right) \implies (\exists n)(X \leq_T 0^{n})$

Where $\phi_i$ is a Turing reduction and $\psi_j$ is an arithmetic formula with one free variable.

We build $A$ via sequence of perfect trees $T_n$ with $T_{n+1}$ a subtree of $T_n$ as in the construction of a minimal arithmetic degree to meet the above requirements with $\lvert T_n(0) \rvert \geq n$. We ensure that each $T_n$ is arithmetic and is uniformly computable from $0^{\omega + 1}$ and set $T_{-1} = 2^{< \omega}$

To meet $P_e$ we simply insist that $T_e(0)$ disagrees with $\phi_i(0^{n})$. Given an index for $T_{e - 1}$ we can find an index for a subtree $\hat{T}_e$ with this property and with $\lvert T_n(0) \rvert \geq e$ computably in $0^{\omega}$.

To meet $R_{i,j}$ with $e = \langle i,j \rangle$ now ask if there is a string $\tau$ on $\hat{T}_e$ and an $x$ such that $\tau \Vdash \psi_j(x) = k \land \phi_i(0^{n};x) = 1 -x$ (here we are using local forcing on $\hat{T}_e$). If so we let $T_e$ be the subtree of $\hat{T}_e$ above $\tau$ and otherwise let it be $\hat{T}_e$. Note that this whole construction can be carried out computably in $0^{\omega +1}$

It's obvious that the requirements $P_e$ are satisfied and we've constructed $A \leq_T 0^{\omega+1}$ so $A$ clearly satisfies $0 <_a A \leq_A 0^{\omega}$. If $R_{i,j}$ is also satisfied then we are done since any $\widehat{A} \leq_a A$ with $\widehat{A} \leq_T 0^{\omega}$ would necessarily be arithmetic and therefore $A$ couldn't be arithmetic in $\widehat{A}$.

So suppose that for some $i,j,n, X$ we have a $\Sigma^0_n$ formula $\psi_j$ defining $X$ from $A$ and $X = \phi_i(0^{\omega})$. Now since $\hat{T}_{\langle i,j \rangle}$ is arithmetic we can find an arithmetic path $Z$ through $\hat{T}_{\langle i,j \rangle}$ that's $n$-generic under local forcing. Thus, the set defined by $\psi_j$ applied to $Z$ is also arithmetic and by construction must be equal to $\phi_i(0^{\omega}) = X$ (or we'd have found a disagreement). Hence, $X$ is arithmetic (i.e. computable from $0^{m}$ for some $m$).

Ok, I'm pretty sure the answer is no to the basic question (still not sure about the generalization). We build a set $A$ to satisfy the following conditions.

$P_e: A \neq \phi_i(0^{n})$ $R_{i,j}: X = \phi_i(0^\omega) \land (\forall i)\left(X(i) = A \models \psi_j(i)\right) \implies (\exists n)(X \leq_T 0^{n})$

Where $\phi_i$ is a Turing reduction and $\psi_j$ is an arithmetic formula with one free variable.

We build $A$ via sequence of perfect trees $T_n$ with $T_{n+1}$ a subtree of $T_n$ as in the construction of a minimal arithmetic degree to meet the above requirements with $\lvert T_n(0) \rvert \geq n$. We ensure that each $T_n$ is arithmetic and is uniformly computable from $0^{\omega + 1}$ and set $T_{-1} = 2^{< \omega}$

To meet $P_e$ we simply insist that $T_e(0)$ disagrees with $\phi_i(0^{n})$. Given an index for $T_{e - 1}$ we can find an index for a subtree $\hat{T}_e$ with this property and with $\lvert T_n(0) \rvert \geq e$ computably in $0^{\omega}$.

To meet $R_{i,j}$ with $e = \langle i,j \rangle$ now ask if there is a string $\tau$ on $\hat{T}_e$ and an $x$ such that $\tau \Vdash \psi_j(x) = k \land \phi_i(0^{n};x) = 1 -x$ (here we are using local forcing on $\hat{T}_e$). If so we let $T_e$ be the subtree of $\hat{T}_e$ above $\tau$ and otherwise let it be $\hat{T}_e$. Note that this whole construction can be carried out computably in $0^{\omega +1}$

It's obvious that the requirements $P_e$ are satisfied and we've constructed $A \leq_T 0^{\omega+1}$ so $A$ clearly satisfies $0 <_a A \leq_A 0^{\omega}$. If $R_{i,j}$ is also satisfied then we are done since any $\widehat{A} \leq_a A$ with $\widehat{A} \leq_T 0^{\omega}$ would necessarily be arithmetic and therefore $A$ couldn't be arithmetic in $\widehat{A}$.

So suppose that for some $i,j,n, X$ we have a $\Sigma^0_n$ formula $\psi_j$ defining $X$ from $A$ and $X = \phi_i(0^{\omega})$. Now since $\hat{T}_{\langle i,j \rangle}$ is arithmetic we can find an arithmetic path $Z$ through $\hat{T}_{\langle i,j \rangle}$ that's $n$-generic under local forcing. Thus, the set defined by $\psi_j$ applied to $Z$ is also arithmetic and by construction must be equal to $\phi_i(0^{\omega}) = X$ (or we'd have found a disagreement). Hence, $X$ is arithmetic (i.e. computable from $0^{m}$ for some $m$).