The sum $\sum_\mu \chi^\lambda_\mu$ over partitions $\mu$ of $n$ is the multiplicity of the irreducible $\chi^\lambda$ in the character afforded by $\mathfrak S_n$ acting on itself by conjugation. If $\psi$ is the character for conjugation, then $\psi(g)$ is the size of the centralizer of $g$, so
$$\langle \chi^\lambda,\psi\rangle =\sum_{\mu\vdash n} \frac{\chi^\lambda_\mu z_\mu}{z_\mu},$$ where $z_\mu$ is the size of the centralizer of any permutation with cycle type $\mu$. So your sum is counting how many irreducibles the conjugation representation breaks up into.

I should mention that the individual row sums $\sum_{\mu\vdash n} \chi^\lambda_\mu$ are not completely understood. It is already non-trivial to prove that they are greater than $0$ (this should be in "An explicit model for the complex representations of $S_n$" by Inglis, Richardson and Saxl.) This is unlike the situation for the group acting on itself by left multiplication. However, when you take your double sum, then there is a good combinatorial description. In fact, summing down any column we know:
$$\sum_\lambda \chi^\lambda_\mu = \# \{g\in \mathfrak S_n : g^2=h\}$$
for any fixed $h\in\mathfrak S_n$ with cycle type $\mu$. This works because each irreducible character of $\mathfrak S_n$ is the character of a real representation. In general, there is the work of Frobenius--Schur. (This is what Lucia was already quickly and correctly observing. Indeed, the sum of degrees will dominate.)

For your second question, it comes from the paper linked to by Lucia in your previous post. Miller proved the statement you are observing (see Theorem 2 and its proof, particularly the last few lines). The number of even entries plus the number of odd entries (denoted $E_n$ and $O_n$) is congruent to the number of partitions $p(n)$, and $E_n$ is always even. So $O_n\equiv p(n)$ mod 2.