Compute $ \sum\limits_{m=1}^{\infty} \sum\limits_{n=1}^{\infty} \sum\limits_{p=1}^{\infty}\frac{(-1)^{m+n+p}}{m+n+p}$

Here is a simple lemma:

Let $(u_n)_{n\geqslant1}$ denote a decreasing sequence of positive functions defined on $(0,1)$, which converges pointwise to zero and such that $u_1$ is integrable on $(0,1)$. Then, $$ \sum\limits_{n=1}^{+\infty}(-1)^n\int_0^1u_n(s)\,\mathrm ds=\int_0^1u(s)\,\mathrm ds,\qquad u(s)=\sum\limits_{n=1}^{+\infty}(-1)^nu_n(s). $$

Now, let us consider the multiple series the OP is interested in. One sees readily that it does not converge absolutely hence the idea is to apply the lemma three times.

• First, fix $n$ and $m$ and, for every $p\geqslant1$, consider $u_p(s)=s^{m+n+p-1}$. Then $u(s)=-\dfrac{s^{m+n}}{1+s}$ hence the lemma yields $$ \sum\limits_{p=1}^{+\infty}\frac{(-1)^{m+n+p}}{m+n+p}=(-1)^{m+n}\sum\limits_{p=1}^{+\infty}(-1)^{p}\int_0^1u_p(s)\,\mathrm ds=(-1)^{m+n+1}\int_0^1\frac{s^{m+n}}{1+s}\,\mathrm ds. $$
• Second, fix $m$ and, for every $n\geqslant1$, consider $u_n(s)=\dfrac{s^{m+n}}{1+s}$. Then $u(s)=-\dfrac{s^{m+1}}{(1+s)^2}$ hence the lemma yields $$ \sum\limits_{n=1}^{+\infty}(-1)^{m+n+1}\int_0^1\frac{s^{m+n}}{1+s}\,\mathrm ds=(-1)^m\int_0^1\frac{s^{m+1}}{(1+s)^2}\,\mathrm ds $$
• Third and finally, for every $m\geqslant1$, consider $u_m(s)=\dfrac{s^{m+1}}{(1+s)^2}$. Then $u(s)=-\dfrac{s^{2}}{(1+s)^3}$ hence the lemma yields $$ \sum\limits_{m=1}^{+\infty}(-1)^m\int_0^1\frac{s^{m+1}}{(1+s)^2}\,\mathrm ds=-\int_0^1\frac{s^{2}}{(1+s)^3}\,\mathrm ds. $$

Thus, the triple series the OP is interested in converges and the value $S_3$ of the sum is $$ \color{red}{S_3=-\int_0^1\frac{s^{2}}{(1+s)^3}\,\mathrm ds}=-\int_1^2\frac{s^{2}-2s+1}{s^3}\,\mathrm ds=-\left[\log(s)+\frac2s-\frac1{2s^2}\right]_1^2, $$ that is, $\color{red}{S_3=-\log(2)+\frac58}=-0.06814718\ldots$

The technique above shows more generally that, for every $k\geqslant1$, the analogous series over $k$ indices converges and that the value of its sum is $$ S_k=(-1)^k\int_0^1\frac{s^{k-1}}{(1+s)^k}\,\mathrm ds=(-1)^k\left(\log(2)+\sum_{i=1}^{k-1}(-1)^i{k-1\choose i}\frac1i(1-2^{-i})\right). $$

This didn' fit in a comment $$ \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \left(\sum_{p=1}^{\infty}\frac{(-1)^{(m+n)+p}}{(m+n)+p}\pm(-1)^{m+n}\sum_{k=1}^{m+n}\frac{(-1)^{k}}{k}\right)\\ =\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \left((-1)^{(m+n)}\sum_{p=1}^{\infty}\frac{(-1)^{p}}{(m+n)+p}\pm(-1)^{m+n}\sum_{k=1}^{m+n}\frac{(-1)^{k}}{k}\right)\\ =\sum_{m=1}^{\infty} \sum_{n=1}^{\infty}\left( \color{red}{ (-1)^{(m+n)}\sum_{p=1}^{\infty}\frac{(-1)^{p}}{(m+n)+p}+(-1)^{m+n}\sum_{k=1}^{m+n}\frac{(-1)^{k}}{k}}-(-1)^{m+n}\sum_{k=1}^{m+n}\frac{(-1)^{k}}{k}\right)\\ =\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \left( \color{red}{ (-1)^{(m+n)}\sum_{p=1}^{\infty}\frac{(-1)^{p}}{p}}-(-1)^{m+n}\sum_{k=1}^{m+n}\frac{(-1)^{k}}{k}\right)\\ =\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \color{red}{ (-1)^{(m+n)}\log(2)}-\Phi_{\text{Lerch}}(-1, 1, 1+n+m)+(-1)^{m+n}\log(2)\\ =\underbrace{\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} (-1)^{(m+n)}2\log(2)}_{=0?}-\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \Phi_{\text{Lerch}}(-1, 1, 1+n+m)\\ =-\sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \Phi_{\text{Lerch}}(-1, 1, 1+n+m)\;, $$ and this is where I give up for now. W|A can do some examples, that make me believe, that this doesn't converge...

Ref's: $-(-1)^{m+n}\sum_{k=1}^{m+n}\frac{(-1)^{k}}{k}=-\Phi_{\text{Lerch}}(-1, 1, 1+n+m)+(-1)^{m+n}\log(2)$