How many components of the a rank five tensor on $\mathbb{R}^5$ which is antisymetric under exchange of any pair of indices are independent?
If we write the tensor $E_{i_1i_2i_3i_4i_5}$ then there are $5!$ permutations of the indices which result in a none zero value. I'm not sure how I then account for the degeneracy due to the negative sign picked up in an exchange of indices.
What you seem to be calling rank is what is usually called degree or order. The term rank is usually reserved for another quantity related to a tensor.
If $E$ is an order five tensor on $\mathbb{R}^5$, then for each choice of $i_1, i_2, i_3, i_4, i_5 \in \{1, 2, 3, 4, 5\}$, we have a corresponding component $E_{i_1i_2i_3i_4i_5}$ which is a real number. However, if $E$ is antisymmetric in each of its indices (we usually just say that $E$ is antisymmetric or skew-symmetric), then if we swap two indices, we multiply the component by $-1$. So, for example, we see that $E_{13452} = -E_{13425}$, and hence one of these components determines the other. By repeatedly using the skew-symmetry, I claim that $E_{i_1i_2i_3i_4i_5}$ is determined by the components with the indices $i_1, i_2, i_3, i_4, i_5$ in increasing order. For example,
$$E_{13452} = - E_{13425} = E_{13245} = -E_{12345},$$
so $E_{12345}$ determines $E_{13452}$. What about if two indices are the same? Then I claim the resulting component is zero. For example, $E_{11345} = -E_{11345}$ by interchanging the first two indices. But $E_{11345}$ is a real number; the only real number which is equal to its negative is zero, so $E_{11345} = 0$. So we see that all of the components are determined by the components $E_{i_1i_2i_3i_4i_5}$ where the indices are strictly increasing (i.e. $i_1 < i_2 < i_3 < i_4 < i_5$). As $i_1, i_2, i_3, i_4, i_5 \in \{1, 2, 3, 4, 5\}$, there is only one such component, namely $E_{12345}$.
In summary, an antisymmetric order $5$ tensor on $\mathbb{R}^5$ has one independent component. In particular, such a tensor is uniquely determined by a component with distinct indices.
By a similar argument, we see that the number of independent components of an antisymmetric order $k$ tensor on $\mathbb{R}^n$ is $\binom{n}{k}$.