Number of edge induced subgraphs with given vertex parity
For simplicity, let's restrict ourselves to $k$-regular graphs.

The first part of your update is correct.  Changing notation slightly from what you have introduced, let me define $n_e$ (resp. $n_o$) to be the number even (resp. odd) edge-induced subgraphs of a given graph $G$.  For the (arity $k$) signature $f = [1, -1, -1, \dotsc, -1]$, the Holant problem Holant($f$) does indeed correspond to the problem "Given a graph, compute $n_e - n_o$."  To see this, fix a subset of edges $E'$ (i.e. the subset of edges assigned 1) and consider a vertex $v$.  If none of the incident edges of $v$ are in $E'$, then $v$ is not in the edge-induced subgraph defined by $E'$ and the constraint at $v$ contributes a factor of 1 to the weight of $E'$.  Otherwise, some incident edge of $v$ is in $E'$, then $v$ is in the edge-induced subgraph defined by $E'$ and the constraint at $v$ contributes a $-1$ to the weight of $E'$.  Summing over the weights of all subsets of the edges gives $n_e - n_o$.  Since $n_e + n_o = 2^{|E(G)|}$, we could solve for $n_e$ and $n_o$ if we could knew $n_e - n_o$.

The second part of your update is incorrect.  The $k$-ary signature $f = [1, -1, -1, \dotsc, -1]$ is not affine for $k \ge 3$.  For a (rather explicit) list of the affine signatures, see page 7 of this paper.  For $k \ge 3$, this signature defines a #P-hard Holant problem. To determine the complexity of this problem, notice that $f$ has the tensor rank 2 decomposition $f = 2 [1, 0]^{\otimes k} - [1,1]^{\otimes k}$. Let $\omega_{2k}$ be a $2k$th primitive root of unity. Then $f = [\sqrt[k]{2}, 0]^{\otimes k} + [\omega_{2k},\omega_{2k}]^{\otimes k}$.  Under a holographic transformation by the inverse of $M = \begin{bmatrix} \sqrt[k]{2} &amp; \omega_{2k} \\ 0 &amp; \omega_{2k} \end{bmatrix}$, we have the bipartite Holant problem $\operatorname{Holant}([\sqrt[k]{4}, \omega_{2k} \sqrt[k]{2}, 2 \omega_{2k}^2] | =_k)$.  This problem is #P-hard by Theorem 22 in this paper, even when also restricting to planar graphs.

For $k \le 2$, $f$ is actually affine, so the problem is tractable, but it is also tractable for a much simpler reason as well.  Any Holant problem using only constraints of arity at most 2 is tractable using matrix product and trace.