What you are doing amounts to computing a tensor product and decomposing it into irreducible components. There is a standard way of doing this with <PERSON>-Richardson rule, Young symmetrizers etc. and there are nice pictures, <PERSON> diagrams (http://en.wikipedia.org/wiki/Young_tableau), that help to visualize different types of symmetries. In many cases it is sufficient to use Young diagrams, so there is no need to write indices directly. For example, $\delta$ is a rank two symmetric, it is depicted by a diagram made of two boxes in a row, we can say it is $(2,0,0,...)$ with the meaning that all other rows are of zero length. Your $f$ is a rank three antisymmetric, it is depicted by a diagrams with three boxes in a column, $(1,1,1,0,...)$. Then $\delta \otimes f$ contains two irreducible components, $(3,1,1,0,...)$ and $(2,1,1,1)$. Each box corresponds to one index and different arrangements of the same number of boxes are known to correspond to different types of tensors one can have with the same number of indices. You can look in Hamermesh or Fulton&Harris.
There are several notations that can be of use for you and are actually used by people. 976
First of all it is convenient to denote all indices that belong to the group of indices in which a tensor is symmetric or antisymmetric by the same letter. For example $f^{uuu}$ or $f^{u[3]}$ instead of your $f^{abc}$ and $\delta^{aa}$ or $\delta^{a(2)}$ for your $\delta^{ab}$ and I used round(square) brackets to indicate the number of indices and whether they are symmetric or antisymmetric. But this works for rather simple types of symmetries, like the one you need.
In the case of $T^{a(n)|u[m]}$ that you gave ($n$ symmetric indices and $m$ antisymmetric) there are still only two irreducible components one is given by $T^{a(s-1)u|u[m]}$ and I assume antisymmetrization over all $u$ indices. Since it is already antisymmetric in the last $u[m]$ this requires $m+1$ terms. The second irreducible components is given by $t^{a(n)|au[m-1]}$ and it requires $n+1$ symmetric permutations. This is just a shorthand notation to same time.
The symmetrization operator you defined is strange and I cannot see that you followed your own recipe in the first formula, for example the 8th term $\delta^{cd}f^{abe}$ must be accompanied by $-\delta^{dc}f^{abe}$, this is another permutation that belongs to $5!$ and there is a sign needed according to your procedure. But the two just cancel each other. Actually, this is always true if you try to move antisymmetric indices to a tensor that is symmetric and vice-verse. Hope this is helpful. 945