= Solution
Under an <orthogonal matrix> $R$, the <Levi-Civita symbol> satisfies
$$
R_{ia}R_{jb}R_{mc}\epsilon_{abc}=(\det R)\epsilon_{ijm}.
$$
Use this twice, and use $R_{mc}R_{mf}=\delta_{cf}$ for the contracted index. It follows that
$$
R_{ia}R_{jb}R_{kc}R_{ld}\epsilon_{abe}\epsilon_{cde}
=(\det R)^2\epsilon_{ijm}\epsilon_{klm}=T_{ijkl}.
$$
Therefore $\boxed{T'_{ijkl}=R_{ia}R_{jb}R_{kc}R_{ld}T_{abcd}=T_{ijkl}}$: it is a fourth-rank <isotropic tensor>. Although a single alternating symbol is a <pseudotensor> under reflections, the two determinant factors multiply to one, so this product is an ordinary <tensor> even for improper orthogonal transformations.
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