As
rand al’thor points out
in the solution built upon here,
there must be a way to formulate a
$\small 3$ with only
two $\small 0 \kern1mu$s.
How promising that...
...$~~ 3 = \sqrt{10-1 \, \small\raise.3ex\strut} ~$...
...uses only two $\small 1$s.
And how convenient that
$\small \surd$ and
$ \small \% $
can combine to scale
$\small 1 $ by any positive power of
$ \small 0.1 \,$.
 
So (with ever fewer operations,
at 18 15 14) ...
$\require{begingroup}\begingroup{} \def \@ #1{{ \sqrt { #1 \, \scriptsize\raise.4ex\strut } }}{} \def \sp {{ \kern 3mu \% }}{} \def \fpp {{ \small \kern 2mu \% \kern1mu \% }}{} \def \fp {{ \small \kern 2mu \% \! \raise.4ex\strut }}{} \def \p {{ \scriptsize \kern 1mu \% }}{} \def \E { ~~ \equiv ~~ }{} \def \= { ~~ = ~~ }{} \def \- { \, - \, }{} \kern 5em \small\begin{align}{}\normalsize { 0! + \@{\@ { 0!\fp }\sp \- 0! \fpp } \over 0!\fp }{} &\= { 1 + \@{ \@ { 1\p \! } \sp \- 1 \p\p } \over 1 \p }{}\\[3ex] &\= { 1 + \@{ \@{ .01 \! } \sp \- 1 \p\p } \over 1 \p }{}\\[2ex] &\= { 1 + \@{ .1 \p \- 1 \p\p } \over 1 \p }{}\\[2ex] &\= { 1 + \@{ .001 - .0001 } \over .01 }{}\\[2ex] &\= { 1 + \@{ .0009 } \over .01 }{}\\[2ex] &\= { 1 + .03 \over .01 }{}\\[2ex] &\E \normalsize 103{}\end{align}\endgroup$
Initial solution.
More operations but also more symmetry.
$\begingroup \displaystyle \kern6em{} \def \@ #1{{ \sqrt { #1 \, } }}{} \def \fp {{ \small \kern 2mu \% \raise.4ex\strut }}{} \def \p {{ \small \kern 1mu \% }}{} { 0! + \@{ \@{ 0!\fp\p\p } \, - \, \@{ 0!\fp\p\p\p } } \over 0!\fp }{}\endgroup$
Backpuzzle.
How about $\boldsymbol{.4}$ with only 3 (three) $\small 0 \kern1mu$s
and just 11 operations?
$\begingroup{} \def \@ #1{{ \sqrt { #1 \scriptsize\raise.4ex\strut } }}{} \def \fp {{ \small \kern 2mu \% \scriptsize\raise.9ex\strut }}{} \def \p {{ \scriptsize \kern 1mu \% }}{} \kern 5em \small\begin{align}{}\normalsize \sqrt { 0!\fp } + \sqrt {\sqrt{ 0!\fp } - 0!\fp \, }{} & ~~ = ~~\@ { 1\p } + \@ { \@ { 1\p } - 1\p \, }{}\\[2ex] & ~~ = ~~ \@ { .01 } + \@ { \@ { .01 } - .01 \, }{}\\[2ex] & ~~ = ~~ \@ { .01 } + \@ { .1 - .01 \, }{}\\[2ex] & ~~ = ~~ \@{ .01 } + \@ { .09 }{}\\[2ex] & ~~ = ~~ .1 + .3{}\\[2ex] & ~~\equiv ~~ \normalsize .4{}\end{align}\endgroup$
$\large($And a ${0! \over 0!\%\%\%}$
more thanks to rand al’thor for fixing up
$\scriptsize\sqrt{\%\%\%\%\,\tiny\raise1ex\strut}$
from the initial solution!
$\large)$
0 / 0! = 103
, right? $\endgroup$