There are no other examples of degree $12$ besides the one you found. Moreover, there are no examples of degrees $10$ or $14$, and in fact the only additional degrees up to $32$ in which there exist examples are $16$, $28$, and $32$, where the groups in the latter cases are precisely (in Klueners/GAP/MAGMA notation) 16T447, 16T777, 16T1079, 16T80, 16T1329, 16T1508, 16T1653, 16T1654, 16T1753, 16T1840, 16T1906, 28T165, 32T35272, 32T397065, 32T2795174. Of these, only the first three degree-$16$ groups and the first two degree-$32$ groups are solvable; those degree-$16$ groups are the unique solvable transitive degree-$16$ groups of orders $240$, $480$, and $960$, and the solvable degree-$32$ groups are the unique transitive degree-$32$ groups of orders $992$ and $4960$. The biggest nonsolvable degree-$16$ group on this list is $\text{AGL}_4(2)$, the degree-$28$ group is $\text{P}\Gamma\text{L}_2(8)$, and the
nonsolvable degree-$32$ group is $\text{AGL}_5(2)$.
More generally, let $f(X)$ be a separable irreducible polynomial over a field $K$, and suppose that $n:=\text{deg}(f)$ is even. Suppose in addition that, if $\{C_1,C_2,\dots,C_{n/2}\}$ and $\{D_1,D_2,\dots,D_{n/2}\}$ are any two distinct partitions of the roots of $f(X)$ into $n/2$ two-element sets, then
$$
\sum_{i=1}^{n/2} \prod_{x\in C_i} x \ne \sum_{i=1}^{n/2}\prod_{x\in D_i} x.
$$
(Note that this condition will hold for any "randomly selected" polynomial having any prescribed Galois group.) Under these hypotheses, if $G$ denotes the Galois group of $f(X)$ over $K$, then the following are equivalent:
- there is an ordering $x_1,\dots,x_n$ of the roots of $f(X)$ such that $x_1x_2+x_3x_4+\dots+x_{n-1}x_n$ has degree $n-1$ over $K$.
- there is an index-$(n-1)$ subgroup $H$ of $G$, and a pair of distinct roots $(x_1,x_2)$ of $f(X)$, such that every element of $H$ maps $\{x_1,x_2\}$ to either itself or to a disjoint set of roots, but every subgroup of $G$ which properly contains $H$ must contain an element which maps $\{x_1,x_2\}$ to a set having exactly one element in common with $\{x_1,x_2\}$.
This can be used to test your condition for specific values $n$, and perhaps group-theoretic results can be used to prove theorems for infinitely many $n$. For instance, I will show later that $G=\text{AGL}_d(2)$ has the required property for $n=2^d$.
To see this equivalence, first assume that the first condition holds; then we can let $H$ be the set of elements in $G$ which fix $y:=x_1x_2+x_3x_4+\dots+x_{n-1}x_n$. This $H$ will be an index-$(n-1)$ subgroup of $G$, and every element of $H$ permutes the collection of 2-element sets $\{\{x_1,x_2\}, \{x_3,x_4\}, ..., \{x_{n-1},x_n\}\}$. Thus, each element of $H$ maps $\{x_1,x_2\}$ to some $\{x_i,x_{i+1}\}$ with $i$ odd. Moreover, since $f(X)$ is irreducible, $G$ acts transitively on $\{x_1,\dots,x_n\}$, so $[G:G_{x_1}]=n$; but since $[G:H]=n-1$ is coprime to $n$, we have $[G:G_{x_1}\cap H]=n(n-1)$ and thus $[H:G_{x_1}\cap H]=n$, so $H$ is transitive. Any subgroup $J$ of $G$ which properly contains $H$ must contain an element $j$ which does not fix $y$, so this element must not permute the collection $\{\{x_1,x_2\}, \{x_3,x_4\}, ..., \{x_{n-1},x_n\}\}$; by multiplying $j$ on both sides by suitable elements of $H$, we obtain an element of $J$ which maps $\{x_1,x_2\}$ to a set having one element in common with $\{x_1,x_2\}$.
Conversely, assume that the second condition holds. As above, $H$ is transitive. It follows that the images of $\{x_1,x_2\}$ under $H$ consist of $n/2$ pairwise disjoint two-element sets, which we can write as $\{x_1,x_2\}, \{x_3,x_4\}, ..., \{x_{n-1},x_n\}$. Then $y:=x_1x_2+x_3x_4+...+x_{n-1}x_n$ is fixed by $H$, but not by any larger subgroup of $G$, so $[K(y):K]=n-1$. This completes the proof of the equivalence.
Added later: here is a proof that $G:=\text{AGL}_d(2)$ has the required property for $n=2^d$. The group $G$ acts on the $\mathbf{F}_2$-vector space $V:=(\mathbf{F}_2)^d$, and $\text{GL}_d(2)$ consists of the elements fixing $0$. Pick some nonzero $c\in V$, and let $J$ be the stabilizer of $c$ in $G$. Let $H$ be the subgroup of $G$ generated by $J$ and the translations $x\mapsto x+u$. I claim that $H$ and $\{0,c\}$ have the required properties. Note that $\text{GL}_d(2)$ acts transitively on $V\setminus\{0\}$, so $[\text{GL}_d(2):J]=2^d-1$. Since the group of translations has order $2^d$, and is normalized by $\text{GL}_d(2)$ (and hence by $J$), we have $\#H=2^d\cdot\#J$ so $[G:H]=2^d-1$. Every translation maps $\{0,c\}$ to a set $\{b,b+c\}$, and any such set either equals $\{0,c\}$ or is disjoint from $\{0,c\}$. Finally, let $K$ be a subgroup of $G$ which properly contains $H$. Since $K$ contains all the translations, and every element of $G$ is a translation times an element of $\text{GL}_d(2)$, it follows that $K$ contains an element of $\text{GL}_d(2)$ which is not in $H$. Any such element maps $c$ to some $b\notin\{0,c\}$, and hence maps $\{0,c\}$ to the set $\{0,b\}$ which has exactly one element in common with $\{0,c\}$.
