The $T[t]=-t$ does not make sense formally, even if I understand what you are trying to say. Time is not an operator, a mere real parameter.
The mistake is in your original calculation. Remember that $T[x]=x$ and $T[p]=-p$ as it should (by definition), consistent with classical expectations. This gives:
$$
T[a] = T[\frac{x+ip}{\sqrt 2}]\\
= \frac{T[x]-iT[p]}{\sqrt 2}\\
= \frac{x+ip}{\sqrt 2}\\
= a
$$
(first use anti-linearity and the effect on $x,p$, both of which cancel out). This gives the expected $T[n] = T[a^\dagger]T[a] = n$.
In the Schrödinger picture, the time evolution of the kets is governed by the equation:
$$
\frac{d}{dt}|\psi(t)\rangle = -iH|\psi(t)\rangle
$$
Since $H = \frac{1}{2}(p^2+x^2) = n+\frac{1}{2}$ is reversible, writing $|\psi(t)\rangle$ (resp $|\phi(t)\rangle$) the evolution of the ket $|\psi\rangle$ (resp $|\phi\rangle$) at $t=0$ with $|\phi\rangle = T|\psi\rangle$, you have:
$$
T|\psi(t)\rangle = |\phi(-t)\rangle
$$
(same initial condition and same equation of motion), in particular if $|\psi\rangle=|\phi\rangle$:
$$
T|\psi(t)\rangle = |\psi(-t)\rangle
$$
Applying this to your example, I think you forgot to mention that $|\psi(0)\rangle$ is time reversible (important assumption that is not always verified, for example a non-zero momentum eigenfunctions), in which case you indeed find:
$$
Ta^\dagger|\psi(t)\rangle = a^\dagger|\psi(-t)\rangle
$$
In the Heisenberg picture, the time dependence of an observable $O(t)$ with initial value $O$ is governed by the equation:
$$
\dot O(t) = -i[O(t),H]
$$
Similarily, considering the time evolved $\tilde O(t)$ with initial value $\tilde O = T[O]$, you'd get:
$$
T[O(t)] = \tilde O(-t)
$$
in particular, if $O = \tilde O$:
$$
T[O(t)] = O(-t)
$$
For example, applying this to $a$, with $a(t) = e^{-it}a$ (from solving Heisenberg's equations), you indeed get (not forgetting anti-linearity):
$$
T[a(t)] = a(-t) (= e^{it}a)
$$
Hope this helps, tell me if something's not clear.
Edit (answers to comment)
Yes with usual composition. Note that since you’re conjugating by an anti-linear operator, the resulting operator is linear.
In your example, I meant $T|\psi(0)\rangle= |\psi(0)\rangle$.
No it is not. Even classically you can see the problem. With $x(t)$ evolving under a time reversible force, you have $x(-t)=\tilde x(t)$ with $\tilde x$ evolving under the same equations of motion, same initial position but opposing initial velocity. In this example, you can see it explicitly. Since $T|x\rangle= |x\rangle$, you have:
$$
T|p\rangle=T\int dx e^{ipx}|x\rangle \\
=\int dx e^{-ipx}T|x\rangle \\
=\int dx e^{-ipx}|x\rangle \\
= |-p\rangle
$$
(using anti-linearity and we obtain the expected result)
Yes sorry typo, corrected it.
As I stated above, your rule $t\to-t$ is false. $t$ is a real parameter, and $T$ is still real linear. This is usually guides you for what to expect when applying $T$ to a time dependent object, but you cannot use it for an actual computation