I know you were asking for a reference, and there may be better approaches, but just to offer one proof of your statement based on the Murnaghan–Nakayama formula. Assume $m>0$ then any skew tableaux $\lambda/\mu$ has shape $(a,1^b)$ with $a+b=n$. This means $a$ boxes in the first row and $b$ boxes in the first column but the row and column are disconnected. We want to evaluate the sum

$$
\Lambda_n = \sum_{a=0}^n \chi_{(a,1^{n-a})}
$$

If $1 \leqslant k \leqslant n$ then the skew partition $(a,1^b)$ has at most two $k$-hooks, one of leg length 0 if $k \leqslant a$ and one of leg length $k-1$ if $k \leqslant b$. Now assume $\nu\vdash n$ is a partition with part equal to $k$ and let $\hat{\nu} \vdash n-k$ be the partition obtained by removing this part. By the Murnaghan–Nakayama formula for skew partitions we have

\begin{align*}
\Lambda_n(\nu) &= \sum_{a=0}^{k-1}\chi_{(a,1^{n-a})}(\nu) + \sum_{a=k}^{n-k}\chi_{(a,1^{n-a})}(\nu) + \sum_{a=n-k+1}^n\chi_{(a,1^{n-a})}(\nu)\\
&= \sum_{a=0}^{k-1}(-1)^{k-1}\chi_{(a,1^{n-a-k})}(\hat{\nu}) + \sum_{a=k}^{n-k}(\chi_{(a-k,1^{n-a})}(\hat{\nu}) + (-1)^{k-1}\chi_{(a,1^{n-a-k})}(\hat{\nu})) + \sum_{a=n-k+1}^n\chi_{(a-k,1^{n-a})}(\hat{\nu})\\
&= \sum_{a=0}^{n-k}\chi_{(a,1^{n-k-a})}(\hat{\nu}) + (-1)^{k-1}\sum_{a=0}^{n-k}\chi_{(a,1^{n-k-a})}(\hat{\nu})\\
&= (1+(-1)^{k-1})\Lambda_{n-k}(\hat{\nu}).
\end{align*}

Hence if $k$ is even then $\Lambda_n(\nu) = 0$ and an easy induction shows that $\Lambda_n(\nu) = 2^{\ell(\nu)}$ if $\nu \in \mathrm{Odd}(n)$.

Maybe I would just add that the same argument essentially works when $m=0$. For this case let $\tilde{\chi}_{(a,1^{n-a})}$ be the irreducible character labelled by a hook partition and let $\tilde{\Lambda}_n = \sum_{a=1}^n \tilde{\chi}_{(a,1^{n-a})}$. The calculation above goes through in this case except when $\nu = (n)$ in which case we have

\begin{equation*}
\tilde{\Lambda}_n(\nu) = \sum_{a=1}^n \chi_{(a,1^{n-a})}(\nu) = \sum_{a=1}^n (-1)^{n-a}
\end{equation*}

which is 0 if $n$ is even and 1 if $n$ is odd. This explains why one gets $\tilde{\Lambda}_n(\nu) = 2^{\ell(\nu)-1}$ when $\nu \in \mathrm{Odd}(n)$.

EDIT: It should also be possible to show that $\Lambda_n = 2\tilde{\Lambda}_n$ directly using the fact that $\chi_{\lambda/\mu} = \sum_{\tau \vdash n} c^{\lambda}_{\mu\tau}\tilde{\chi}_{\tau}$, where $c^{\lambda}_{\mu\tau}$ is the Littlewood–Richardson coefficient.