Here's an argument showing that in the ${\rm II}_1$ case the flip automorphism is never inner.

Let $M$ be a type ${\rm II}_1$ factor and $\tau$ its trace, so that $M \subset L^2(M, \tau)$, and $M$ acts standardly on $L^2(M, \tau)$. Suppose that the flip automorphism is implemented by a unitary $U \in \mathcal U(M \overline \otimes M)$. I'll reach a contradiction by showing that $U$ is orthogonal to every vector in the dense subspace of $L^2(M \overline \otimes M, \tau \otimes \tau)$ spanned by vectors of the form $v \otimes w$ where $v, w \in \mathcal U(M)$.

Fix $v, w \in \mathcal U(M)$, and $\varepsilon > 0$. Take $n \in \mathbb N$ such that $2^{-n} < \varepsilon$, and take a partition of unity $\{ p_k \}_{k = 1}^{2^n} \subset M$ such that each $p_k$ is a projection of trace $2^{-n}$. Then
$$
| \langle U, v \otimes w \rangle | = \left| \sum_{k = 1}^{2^n} (\tau \otimes \tau) ((p_k \otimes 1)(v^* \otimes w^*) U (p_k \otimes 1) ) \right|
$$
$$
\leq \sum_{k = 1}^{2^n} | (\tau \otimes \tau) ((v^* \otimes w^*) U (p_k \otimes vp_kv^*) ) |
$$
$$
\leq \sum_{k = 1}^{2^n} (\tau \otimes \tau)(p_k \otimes vp_kv^*)
$$
$$
= \sum_{k = 1}^{2^n} \tau(p_k)^2 = 2^{-n} < \varepsilon.
$$

**Update**: I've recently come across the 1975 paper of Sakai "Automorphisms and tensor products of operator algebras" where he proves that the flip automorphism for a von Neumann algebra $M$ is inner if and only if $M$ is a type ${\rm I}$ factor. His proof is roughly as follows:

For the type ${\rm II}_1$ case he proceeds as I did above by showing that the unitary $U$ would have to be orthogonal to every vector of the form $v \otimes w$. His argument for this is not as direct as the one above, but the argument I gave above is based on techniques of Popa which came later.

For the type ${\rm II}_\infty$ case he writes $M$ as $N \overline \otimes \mathcal B(\mathcal H)$ where $N$ is a type ${\rm II}_1$ factor and then shows with a simple argument that if the flip automorphism is inner on $M$ then it must also be inner on $N$ which it cannot be by the arguments above.

For the type ${\rm III}$ case he first writes $M$ as $N \overline \otimes \mathcal B(\mathcal H)$ where $N$ is type ${\rm III}$ and countably decomposable. Next he shows that if the flip is inner on $N$ then $N$ has trivial outer automorphism group. Indeed, if $\sigma$ denotes the flip and $\rho \in {\rm Aut}(N)$ then since $\sigma$ is inner, and ${\rm Inn}(M)$ is a normal subgroup, we must have $\tilde \rho = \sigma (\rho^{-1} \otimes {\rm id}) \sigma (\rho \otimes {\rm id})$ is also inner. Restricting $\tilde \rho$ to $N \otimes \mathbb C$ we then see then that there is a unitary $V \in N \overline \otimes N$ such that $(a \otimes 1)V = V(\rho(a) \otimes 1)$ for all $a \in N$. If we then consider the normal conditional expectation $E$ from $N \overline \otimes N$ to $N \otimes \mathbb C$, then there exists some operator $x \in \mathbb C \otimes N$ such that $E(Vx) \not= 0$, and we then have $a E(Vx) = E(Vx) \rho(a)$ for all $a \in N$. By conjugating this formula it then follows easily that $E(Vx)E(Vx)^* \in \mathcal Z(N) = \mathbb C$ and also $E(Vx)^*E(Vx) \in \mathbb C$, hence $E(Vx)$ is a non-zero scalar multiple of a unitary showing that $\rho$ is inner. Tomita-Takesaki theory though gives continuum many outer automorphism of $N$, a contradiction.