[As explained by Michael Pilat][1] you cannot create your own compound operators\* with custom precedence.  (You could conceivably write your own parser as Leonid has worked on, or attempt to coerce the Box form with `CellEvaluationFunction`.)

You can however use an existing operator with the desired precedence.  Looking at [the table][2] `Colon` appears to be a good choice.  The operator is entered with <kbd>Esc</kbd>`:`<kbd>Esc</kbd>.  Example:

    SetAttributes[Colon, HoldAll]
    Colon[f__, x_] := Composition[f][Unevaluated@x]
    
    ListPlot \[Colon] Flatten \[Colon] Log@N@Accumulate@# & /@ Partition[Range@300, 100]

Which appears as, and produces:

![Mathematica graphics](https://i.sstatic.net/zAwzZ.png)

![Mathematica graphics](https://i.sstatic.net/uCsR4.png)

Since raw colon is already used for `Pattern` this may be confusing.  However, if you are willing to edit your `UnicodeFontMapping.tr` file you can assign any symbol you like.  Here I mapped `\[Colon]` to [Klingon][3] A:

![Mathematica graphics](https://i.sstatic.net/Y1SZN.png)

This was done by changing the line starting with `0x2236` in `UnicodeFontMapping.tr`.

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\* Rojo demonstrated that one can create *two-dimensional* compound operators, meaning use of [`SubscriptBox`](http://reference.wolfram.com/language/ref/SubscriptBox.html), [`SuperscriptBox`](http://reference.wolfram.com/language/ref/SuperscriptBox.html), [`OverscriptBox`](http://reference.wolfram.com/language/ref/OverscriptBox.html), etc.  See:

- https://mathematica.stackexchange.com/questions/27081/is-it-possible-to-define-custom-compound-assignment-operators-like-%E2%8A%95-similar-to/39703#39703


  [1]: https://stackoverflow.com/a/5305511/618728
  [2]: http://reference.wolfram.com/mathematica/tutorial/OperatorInputForms.html
  [3]: https://mathematica.stackexchange.com/q/1278/121