After an exhaustive search (by hand) I've found a really good solution for 0 using 14 digits!
$(0!+0!+0!)\times(((0!+0!+0!)!)!-((0!+0!+0!)!+0!))-((0!+0!+0!)!-0!)!$
$=3\times((3!)!-(3!+1))-(3!-1)!$
$=3\times(6!-(6+1))-(6-1)!$
$=3\times(720-7)-5!$
$=3\times(713)-120$
$=2139-120$
$=2019$
Proof for optimisation
To make this proof less convoluted (and to help in future similar problems) this is a dictionary of the construction of all numbers from 1-10. [Number ($i$ count): Expression]
1 (1): $0!$
2 (2): $1+1$
3 (3): $2+1$
4 (4): $3+1$ or $2^2$
5 (4): $3!-1$
6 (3): $3!$
7 (4): $6+1$
8 (5): $2^3$ or $6+2$
9 (5): $3^2$
10 (6): $9+1$ or $2\times5$
Proof that it isn't possible to reach 2019 using two (1-10) number $i$-expressions, so at least more than 8 digits are required
The only way to reach 2019 using two $i$-expressions is to use an expression either of the form $A \times B$, $A^B$ or $A-B$ ($\frac{A}{B}$ and $A+B$ are redundant). The only factors of 2019 are 3 and 673 and no factorial can equal 673 so $A \times B$ is ruled out and $A^B$ is ruled out by default. For $A-B$ to bear a solution, $A \ge 2019$ however no large number $B$ obtainable by a 1-10 number expression exists to constrain possible large $A$.
Following off of that proof, we now know the approach to take to get 2019
$A^B$ and $\frac{A}{B}$ are completely out of the question. From this it is safe to say that either $3\times673=2019$, $A+B$, $A-B$ or $A=2019^B$
On the subject of factorials...
No perfect power factorial exists above $1!$. Also no expression containing only factorials (above $1!$) can equal a prime number due to common factors.
This helps to drastically rule out many expressions
$A!=2019^B$ is out. $673$ is a prime number so the only way to reach it is by $A \pm B$. In the case of $673$, $A \pm B$ is disprovable by exhaustion for 1-10 numbers since nested factorials are disproved. In a more general case, $A=2019^B$ cannot be done if $C$ is added either. For it to even bear results it would have to be $A \pm B = 2019^C$ or $A \times B = 2019^C$. Since $2019^C=3^C673^C$ it clearly isn't possible using nested factorials so this can be disproved by trivial exhaustion. A common factor of $3^C$ can be brought out of $A!\pm B!$ for $A!,B! > 3!$ leaving two other factorials which must sum or difference to $673^C$ which isn't possible because it is many of the same prime. This is now trivially disprovable by exhaustion! Thus this case generally requires at least 12+ digits since more than three (1-10) number $i$-expressions are required. I doubt this will lead anywhere so it is fair to say that it has been disproved! {of course this isn't 100% irrefutable proof since it is possible for $D$ to enter the equation however the numbers (1-10) would be highly restrained and I don't want to write a full paper on this}
On the subject of $3\times673=2019$ we already know that $3$ takes 3 digits so all I have to do is show that $673$ cannot be made in a reasonable amount of digits. We know that $A,B$ expressions that contain only factorials cannot be prime and the other solutions are trivially disprovable by exhaustion ($A \pm B$). This brings it up to 11+ digits. Adding in $C$, the cases to be examined are $A \times B \pm C$, $A-B \times C$ and $A^B-C$. The unlisted cases are either clearly logically wrong, already listed and/or trivially disproved by exhaustion. $A \times B \pm C$ requires a nested factorial somewhere in order for the numbers to be large enough for a solution. If $A$ or $B$ is a factorial $>1!$ then $C$ must be an odd number for the result to be prime. Therefore either all numbers are nested factorials (disproved by previous logic) or only one of them are a nested factorial. Since only one of them can be a nested factorial, this is disprovable by exhaustion (gee this is getting exhausting!). This is extended to $A-B \times C$ (which is now very obvious without proof by "..."). For $A^B-C$, $A$ and $C$ cannot both be nested factorials so logically only $B,C$ should be examined. $A$ must then be odd. Looking at $A^{B!}-C!$, the cases where $A \le C$ - have a common factor thus disproved. $A>C$ restricts nested factorial $C$ to $A$'s upper bound thus is disprovable by "...". All of this proof comfortably disproves $3\times673=2019$ since $673$ requires more than three (1-10) numbers and at worse case scenario (unlike my $A=2019^B$ proof) is +12 digits but should generally require +15 digits.