The case for antisymmetric tensors is not different from other generic tensors. Normally one has
$$
T_{\mu_1\cdots \mu_k} = \eta_{\mu_1\nu_1}\cdots \eta_{\mu_k\nu_k} T^{\nu_1\cdots \nu_k}\,.
$$
If you have a totally antisymmetric tensor this simplifies to
$$
\epsilon_{\mu\nu\rho\lambda} = \det(\eta) \,\epsilon^{\mu\nu\rho\lambda}\,.
$$
And since $\eta = \mathrm{diag}(1,-1,-1,-1)$ you get the minus, the same would be true for the mostly plus metric (in any even dimension). For the electric field one has to be precise with the definition
$$
E^i \equiv F^{i0}\,,\qquad E_i \equiv F_{i0}\,.
$$
And you can see that for lowering the $i$ index you pay the price of a minus sign. The same would be true for a mostly plus metric but the sign comes from the $0$.
Finally the $\epsilon_{ijk}$. If you define
$$
\epsilon_{ijk} \equiv \epsilon_{0ijk}\,,\qquad \epsilon^{ijk} \equiv \epsilon^{0ijk} = \eta^{00}\eta^{i l} \eta^{jm}\eta^{kn} \epsilon_{lmn}\,, \tag{1}\label{1}
$$
then the upper and the lower index version differ in sign. If you instead define
$$
\epsilon^{ijk} = \eta^{i l} \eta^{jm}\eta^{kn} \epsilon_{lmn}\,, \tag{2}\label{2}
$$
then the relative sign depends on whether you use mostly plus $(+)$ or mostly minus $(-)$ signature. This is not an ambiguity. If you use either definition consistently, when you decompose any four-vector expression into a time and a three-dimensional part you'll get the same result.
For example, $B$ can be defined as
$$
\tilde{F}^{\mu\nu} \equiv -\tfrac12 \epsilon^{\mu\nu\rho\lambda} F_{\rho\lambda}\,,\qquad B^i \equiv \tilde{F}^{i0}\,,\qquad B_i \equiv \tilde{F}_{i0}\,.
$$
And it also changes sign as any other (presudo)vector. Its decomposition into time and 3d part is
$$
B^i = \tfrac12 \epsilon^{0ijk} F_{jk} = s \tfrac12 \epsilon^{ijk} F_{jk}\,,\;\;\qquad B_i = \tfrac12 \epsilon_{0ijk} F^{jk} = \tfrac12 \epsilon_{ijk} F^{jk} \,,
$$
where $s=1$ for the definition \eqref{1} and $\eta^{00}$ for the definition \eqref{2}. You can see that, not matter what, $B^i = - B_i$ (in any signature).
Note that if you are in Euclidean space to start with (i.e. you are not in an Euclidean subspace of Minkowski), then the natural metric is the all-plus $\delta_{ij}$, obviously. Therefore the Levi-Civita tensor with upper and lower indices will have the same sign $\epsilon_{i_1\ldots i_n} = \epsilon^{i_1\ldots i_n}$.