It's easy to see where the loop comes from as \Bigr is defined in terms of \Big which is (by default)
\def\Big#1{{\hbox{$\left#1\vbox to11.5\p@{}\right.\n@space$}}}
so defined in terms of \left
It is probably safer just to do a replace/edit on the incoming file
If you wanted to do this in TeX you'd neeed
\let\oldleft\left
\let\oldright\right
and then redefine each of these
\def\big#1{{\hbox{$\left#1\vbox to8.5\p@{}\right.\n@space$}}}
\def\Big#1{{\hbox{$\left#1\vbox to11.5\p@{}\right.\n@space$}}}
\def\bigg#1{{\hbox{$\left#1\vbox to14.5\p@{}\right.\n@space$}}}
\def\Bigg#1{{\hbox{$\left#1\vbox to17.5\p@{}\right.\n@space$}}}
to use \oldleft and \oldright then you would be slightly safer to do
\let\left\Bigl
\let\right\Bigr
in your local redefinition, although there can still be cases that fail, notably as the \left and \right primitives do not scan for arguments the same way as macros like \Bigr so for example
$\left\delimiter"26A30C \frac{a}{b} \right.$
is OK but if you replace \left with \Bigr you get
$\Bigl\delimiter"26A30C \frac{a}{b} \Bigr.$
which produces
! Missing number, treated as zero.
<to be read again>
\vbox
l.11 $\Bigl\delimiter
as the argument to \Bigr is just the token \delimiter not the full delimiter specification.