How does Kronecker $\delta_{ijkl}$ determined?

Let's have symbol $\delta_{ijkl}$. Is it equal to 1 for $i = j = k = l$ and $0$ in the other cases?

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It depends on the context. $\delta_{ijkl}$ could be defined as a generalization of Kronecker delta (then it is -as you wrote- equal to 1 if $i=j=k=\ell$ and 0 otherwise), but it could also denote the Levi-Civita symbol. In 4 dimensions it's defined as
$$\delta_{ijkl} := \frac{(i-j) \cdot (i-k) \cdot (i-l) \cdot (j-k) \cdot (j-l) \cdot (k-l)}{12}$$
where $i,j,k,l \in \{1,2,3,4\}$. If you want to generalize it you can use permutations, so
$$\delta_{ijkl} = \begin{cases} 1 & (i,j,k,l) \, \, \text{is an even permutation} \\ -1 & (i,j,k,l) \, \, \text{is an odd permutation} \\ 0 & \text{otherwise} \end{cases}$$
I met this Kroneker delta when read about a 4-rank stiffness tensor written for cube-symmetric body. It means that it's invariant under rotations at pi/2 angle. It's components can be representated as $$a_{ijkl} = b\delta_{ij}\delta_{kl} + b(\delta_{ik}\delta_{jl} + \delta_{il}\delta_{jk}) + c\delta_{ijkl},$$ where the last summand is invariant under rotations only at angle $\pi k/2$ round an x-,y-,z- axis. I can't prove it without determine a $\delta_{ijkl}$. – John Taylor Dec 19 '12 at 11:18