It's basically the same, it's mainly a matter of implicit identifications.
Namely, in a Hilbert space $H$ there is a natural isomorphism $H\to H^*$ (which is the bra-ket duality), so $H\otimes H\simeq H\otimes H^*$.
Now there is a map $H\otimes H^*\to L(H)$ which corresponds to what you call the outer product.
So for two kets $|\psi\rangle$ and $|\phi\rangle$ in $H$, you can either see $|\psi\rangle\otimes |\phi\rangle$ as an element of $H\otimes H$ and interpret that as a kind of two-particles state, or you can look at its canonical image in $H\otimes H^*$, which the physicists note $|\psi\rangle \langle\phi|$ (which makes sense since $\langle\phi|$ is the element of $H^*$ corresponding to $|\phi\rangle$), and then interpret it as an operator in $L(H)$.
What you wrote is wrong, sorry. If the Hilbert space is $L^2(\mathbb C^n \times \mathbb R, \mathbb C)$, then the relevant integral is $$\int_{\mathbb C^n \times \mathbb R} \overline{\psi(x,t)} \phi(x,t) d^nx dt\tag{1}$$
where (I am interpreting your notations) the functions are complex-valued and $d^nx$ is the Lebesgue measure over $\mathbb C^n$ viewed as $\mathbb R^{2n}$ and $dt$ the one over $\mathbb R$. Presumably your $\mathbb C^n$ should be $\mathbb R^n$.
Instead you use the measure $d^nx$ over the factor factor only.
$$\int_{\mathbb C^n} \overline{\psi(x)} \phi(x) d^nx$$
In this case it does make any sense to wonder whether or not $\partial_t$ is Hermitian or not, since it is not an operator over the linear space $L^2(\mathbb C^n, \mathbb C)$, that is on the functions $\psi=\psi(x)$
(Even dealing with the whole Hilbert space $L^2(\mathbb C^n \times \mathbb R, \mathbb C)$ and the scalar product (1) there would be other issues regarding regularity of functions but the problem arising from your approach is much more relevant.)
In elementary Quantum Mechanics the Hilbert space is $L^2(\mathbb R^n)$ referred to Lebesgue's measure $d^nx$ over $\mathbb R^n$.
Operators are linear maps $A: D(A) \to L^2(\mathbb R^n)$ where $D(A) \subset L^2(\mathbb R^n)$ is a (usually dense) linear subspace.
You are instead considering a function from $\mathbb R$ to $L^2(\mathbb R^n)$ physically representing the temporal evolution of a vector $\psi_0$ at $t=0$
$$\mathbb R \ni t \mapsto \psi_t \in L^2(\mathbb R^n) $$
The derivative you are considering acts on such a family
$$\partial_t \psi_t := \lim_{h \to 0} \frac{\psi_{t+h} -\psi_t}{h} \tag{2}$$
and, if exists, is computed with respect to the topology of $L^2(\mathbb R^n)$.
You see that to compute (2) you need a $t$-parametrized family of vectors.
With a single vector $\psi \in L^2(\mathbb R^n)$, $\partial_t \psi$ would not have any meaning, differently from, say, $A\psi$ -- where $A$ is a true linear operator in $L^2(\mathbb R^n)$ and $\psi \in D(A)$ -- which is meaningful.
Best Answer
$\langle\psi |\phi\rangle\langle\phi|\psi\rangle$ is a 1x1 matrix (scalar) and the trace of a 1x1 matrix is the scalar itself. Therefore,
$$ \text{Tr}(\langle\psi |\phi\rangle\langle\phi|\psi\rangle) = \langle\psi|\phi\rangle \langle\phi|\psi\rangle $$
Or, in english, the trace of a scalar is the scalar itself!
It has nothing to do with the fact that $$ |\psi\rangle\langle\psi|\phi\rangle\langle\phi| \neq \langle\psi|\phi\rangle \langle\phi|\psi\rangle $$ It is, instead, to do with $$ \text{Tr}(AB)=\text{Tr}(BA) $$ (If A is an $m × n$ matrix and B an $n × m$ matrix)
Then, there is nothing wrong with: $$ \begin{align} \text{Tr}(|\psi\rangle\langle\psi |\phi\rangle\langle\phi|) &= \text{Tr}(\langle\psi |\phi\rangle\langle\phi|\psi\rangle) \\&= \langle\psi|\phi\rangle \langle\phi|\psi\rangle \\&= \langle\psi|\phi\rangle \overline{\langle\psi|\phi\rangle} \\&= | \langle\psi|\phi\rangle | ^2 \end{align} $$
Original Answer
Here is my proposed answer:
$$ \begin{align} \text{Tr}(P_\psi P_\phi) &= \text{Tr}(|\psi\rangle\langle\psi| |\phi\rangle\langle\phi|) \\&= \langle\psi|\phi\rangle \text{Tr}(|\psi\rangle\langle\phi|) \\&= \langle\psi|\phi\rangle \langle\phi|\psi\rangle \\&= \langle\psi|\phi\rangle \overline{\langle\psi|\phi\rangle} \\&= | \langle\psi|\phi\rangle | ^2 \end{align} $$ , which coincides with the original answer.
Nevertheless, I still think that the method ilustrated in the question is wrong. Meaning that: $ |\psi\rangle\langle\psi|\phi\rangle\langle\phi| \neq \langle\psi|\phi\rangle \langle\phi|\psi\rangle $ , where the LHS is an operator and the RHS a scalar.By the way, $$ \text{Tr}(|\psi\rangle\langle\phi|) = \langle\phi|\psi\rangle $$ comes from $$ \langle\phi|\psi\rangle = \sum_n \langle\phi|n\rangle\langle n|\psi\rangle $$