$\DeclareMathOperator{Out}{Out} \DeclareMathOperator{Aut}{Aut} \DeclareMathOperator{IA}{IA}$ The key thing that is missing is that the Torelli subgroup of $\Out(F_n)$ is contained in the normal closure of these elements. This is in fact true, and is a very special case of a theorem will appear in my student Jiayi Shen's thesis. Since More generally, her thesis will prove that in most cases, the $\Out(F_n)$-normal closure of a finite subgroup of $\Out(F_n)$ is what you would guess to be (there are a few cases, and more generally she has to assume that $n \geq 3$ and can't handle everything that happens at the prime $2$, though for the special case asked here she has no issues). Since it is not yet posted to the arXiv, I will give the argument here specialized to this particularly easy case.
Instead of $\Out(F_n)$, I will prove it for $\Aut(F_n)$ (which is enough). Let $\IA_n$ be the Torelli subgroup of $\Aut(F_n)$. Let $\{x_1,\ldots,x_n\}$ be the standard basis of $F_n$. For distinct $1 \leq i,j \leq n$, let $C_{ij} \in \IA_n$ be the element that conjugates $x_i$ by $x_j$. A classical theorem of Nielsen says that the $\Aut(F_n)$-normal closure of $C_{12}$ is $\IA_n$, so it is enough to prove that $C_{12}$ is in the indicated subgroup.
One element of the subgroup in question is the map $f\colon F_n \rightarrow F_n$ that inverts $x_1$ and $x_2$ and fixes all the other $x_i$. Let $L_{ij} \in \Aut(F_n)$ be the Nielsen transformation that multiplies $x_i$ on the left by $x_j$. The subgroup in question then contains $$(L_{12} f L_{12}^{-1}) f^{-1},$$ which equals $C_{12}$ by the following calculation:
- First,
$$L_{12} f L_{12}^{-1} f^{-1}(x_1) = L_{12} f L_{12}^{-1}(x_1^{-1}) = L_{12} f (x_1^{-1} x_2) = L_{12}(x_1 x_2^{-1}) = x_2 x_1 x_2^{-1}.$$
- Second, for $i \geq 2$ for an appropriate choice of sign depending on whether $i=2$ or $i \geq 3$ we have
$$L_{12} f L_{12}^{-1} f^{-1}(x_i) = L_{12} f L_{12}^{-1}(x_i^{\pm 1}) = L_{12} f(x_i^{\pm 1}) = L_{12}(x_i) = x_i.$$