Signs always seem to cause problems, and everyone seems to have their own conventions. Brian Conrad spends 11 pages at the beginning of his book 'Grothendieck duality and base change' discussing such issues.
When I recently had cause to use triple complexes, I felt the easiest approach was to define them having commutative squares, and only introduce signs when taking total complexes. The stacks project seems to take the same view. In that respect this answer may not be that helpful to you, as it uses a different convention to Weibel.
We take a triple complex $T_{ijk}$ (with commuting squares), and with differentials $s,t,u$, so
$$ s_{ijk} : T_{ijk} \to T_{i-1jk}, \quad t_{ijk} : T_{ijk} \to T_{ij-1k}, \quad u_{ijk} : T_{ijk} \to T_{ijk-1}. $$
We can then form the double complex $\mathrm{Tot}_{12}T$, having terms $\prod_{i+j=d}T_{ijk}$ and differentials $(s_{ijk}+(-1)^it_{ijk})$ and $(u_{ijk})$. More precisely, it sends $(x_{ijk})\in\prod_{i+j=d}T_{ijk}$ to the element of $\prod_{i+j=d-1}T_{ijk}$ having term $s_{i+1jk}(x_{i+1jk})+(-1)^it_{ij+1k}(x_{ij+1k})$ in position $ijk$.
We can also form the double complex $\mathrm{Tot}_{23}T$, having terms $\prod_{j+k=e}T_{ijk}$, and differentials $(s_{ijk})$ and $(t_{ijk}+(-1)^ju_{ijk})$.
Finally we can directly form the complex $\mathrm{Tot}T$, having terms $\prod_{i+j+k=d}T_{ijk}$ in position $d$, and differential $(s_{ijk}+(-1)^it_{ijk}+(-1)^{i+j}u_{ijk})$.
It is then easy to check that
$$ \mathrm{Tot}\mathrm{Tot}_{12}T = \mathrm{Tot}T = \mathrm{Tot}\mathrm{Tot}_{23}T, $$
where on the left and right we have taken the total complexes of the double complexes.
So far so good.
Suppose now that we have complexes $(X,s)$, $(Y,t)$ and $(Z,u)$, so $s_i:X_i\to X_{i-1}$ etc. The idea is then that we can form the triple complexes $T=\mathrm{Hom}(X\otimes Y,Z)$ and $U=\mathrm{Hom}(X,\mathrm{Hom}(Y,Z))$, and the standard tensor-hom adjunction quickly shows (there are no sign problems at this stage) that $T\cong U$ as triple complexes. It follows that their total complexes are isomorphic, and using the above factorisations we see that
$$ \mathrm{Tot}\mathrm{Tot}_{12}\mathrm{Hom}(X\otimes Y,Z) \cong \mathrm{Tot}\mathrm{Tot}_{23}\mathrm{Hom}(X,\mathrm{Hom}(Y,Z)). $$
We then show that we have isomorphisms of double complexes
$$ \mathrm{Tot}_{12}\mathrm{Hom}(X\otimes Y,Z) \cong \mathrm{Hom}(\mathrm{tot}(X\otimes Y),Z) $$
and
$$ \mathrm{Tot}_{23}\mathrm{Hom}(X,\mathrm{Hom}(Y,Z)) \cong \mathrm{Hom}(X,\mathrm{Tot}\mathrm{Hom}(Y,Z)). $$
Here we have used the small total complex $\mathrm{tot}(X\otimes Y)$, having terms $\bigoplus_{i+j=d}X_i\otimes Y_j$ and differential $p_d=\sum s_i\otimes1+(-1)^i\otimes t_j$.
That's the plan.
The triple complex $T=\mathrm{Hom}(X\otimes Y,Z)$ has terms $\mathrm{Hom}(X_{-i}\otimes Y_{-j},Z_k)$, and differentials $(s_{1-i}\otimes1)^\ast$, $(1\otimes t_{1-j})^\ast$, and $(u_k)_\ast$.
The triple complex $U=\mathrm{Hom}(X,\mathrm{Hom}(Y,Z))$ has terms $\mathrm{Hom}(X_{-i},\mathrm{Hom}(Y_{-j},Z_k)$ and differentials $(s_{1-i})^\ast$, $((t_{1-j})_\ast)^\ast$ and $((u_k)_\ast)_\ast$ (sorry about the notation!)
The isomorphism $T\cong U$ of triple complexes just comes from the tensor-hom adjunctions $\mathrm{Hom}(X_{-i}\otimes Y_{-j},Z_k)\cong\mathrm{Hom}(X_{-i},\mathrm{Hom}(Y_{-j},Z_k))$. That's the first part out of the way.
The double complex $\mathrm{Hom}(\mathrm{tot}(X\otimes Y),Z)$ has terms $\mathrm{Hom}(\bigoplus_{i+j=-d}X_i\otimes Y_j,Z_k)$, and differentials $(p_{1-d})^\ast$ and $(u_k)_\ast$. Using the standard isomorphism
$$ \prod_{i+j=-d}\mathrm{Hom}(X_i\otimes Y_j,Z_k) \cong \mathrm{Hom}(\bigoplus_{i+j=-d}X_i\otimes Y_j,Z_k) $$
the first differential sends a tuple $f_{ij}:X_i\otimes Y_j\to Z_k$ for $i+j=-d$ to the tuple $f_{i-1j}(s_i\otimes1)+(-1)^if_{ij-1}(1\otimes t_j):X_i\otimes Y_j\to Z_k$ for $i+j=1-d$. This is precisely what the first differential for $\mathrm{Tot}_{12}\mathrm{Hom}(X\otimes Y,Z)$ does, though. Thus the standard isomorphisms above determine an isomorphism of double complexes
$$ \mathrm{Tot}_{12}\mathrm{Hom}(X\otimes Y,Z) \cong \mathrm{Hom}(\mathrm{tot}(X\otimes Y),Z) $$
as required.
Similarly, the standard isomorphisms
$$ \prod_{j+k=e}\mathrm{Hom}(X_{-i},\mathrm{Hom}(Y_{-j},Z_k)) \cong \mathrm{Hom}(X_{-i},\prod_{j+k=e}\mathrm{Hom}(Y_{-j},Z_k)) $$
determine an isomorphism of double complexes
$$ \mathrm{Tot}_{23}\mathrm{Hom}(X,\mathrm{Hom}(Y,Z)) \cong \mathrm{Hom}(X,\mathrm{Tot}\mathrm{Hom}(Y,Z)). $$
In particular, the second differential in both cases sends a tuple $g_{jk}:X_{-i}\to\mathrm{Hom}(Y_{-j},Z_k)$ with $j+k=e$ to the tuple $(t_{-j})^\ast g_{j+1k}+(-1)^j(u_{k+1})_\ast g_{jk+1}$ with $j+k=e-1$.
This proves the second part, and so we are done.
Best Answer
There's an isomorphism between the two versions of the total complex induced by $\text{id}:C^{i,j}\to C^{i,j}$ when
and by $-\text{id}:C^{i,j}\to C^{i,j}$ otherwise.
[In the original version of the question, the two proposed differentials were $d+(-1)^ie$ and $d-(-1)^ne$ (rather than $d+(-1)^ie$ and $e-(-1)^nd$. For that case, an isomorphism between the two versions of the total complex is induced by $\text{id}:C^{i,j}\to C^{i,j}$ for $j\equiv 0,3\pmod{4}$ and $-\text{id}:C^{i,j}\to C^{i,j}$ for $j\equiv 1,2\pmod{4}$.]
In fact, for any two reasonable sign conventions (by which I mean that some arrows in the double complex are assigned minus signs in such a way that every $1\times 1$ square has an odd number of minus signs on its edges), there will be an isomorphism given by $\pm\text{id}$ on each $C^{i,j}$. To decide whether to use $+\text{id}$ or $-\text{id}$, use $+\text{id}$ on $C^{0,0}$, count how many minus signs are assigned to the arrows on a path in the double complex from $(0,0)$ to $(i,j)$. The parity of the difference between the two numbers obtained from the two sign conventions will be independent of the path. Use $+\text{id}$ if this parity is even, and $-\text{id}$ if it's odd.