Supposing that the fair die has $n$ sides and is rolled $m$ times we
get for the probability of $k$ distinct outcomes by first principles
the closed form
$$\frac{1}{n^m} \times
m! [z^m] {n\choose k} (\exp(z)-1)^k.$$
Let us verify that this is a probability distribution. We obtain
$$\frac{1}{n^m} \times
m! [z^m]
\sum_{k=0}^n {n\choose k} (\exp(z)-1)^k.$$
Here we have included the value for $k=0$ because it is a constant
with respect to $z$ and hence does not contribute to $[z^m]$ where
$m\ge 1.$ Continuing we find
$$\frac{1}{n^m} \times
m! [z^m] \exp(nz) = 1$$
and the sanity check goes through. Now for the expectation we get
$$\mathrm{E}[N] =
\frac{1}{n^m} \times
m! [z^m]
\sum_{k=1}^n k {n\choose k} (\exp(z)-1)^k
\\ = \frac{n}{n^m} \times
m! [z^m]
\sum_{k=1}^n {n-1\choose k-1} (\exp(z)-1)^k
\\ = \frac{n}{n^m} \times
m! [z^m] (\exp(z)-1)
\sum_{k=0}^{n-1} {n-1\choose k} (\exp(z)-1)^k
\\ = \frac{n}{n^m} \times
m! [z^m] (\exp(z)-1) \exp((n-1)z).$$
This is
$$\frac{n}{n^m} \times
(n^m - (n-1)^m) = n \left(1-\left(1-\frac{1}{n}\right)^m\right).$$
This goes to $n$ in $m$ as it ought to. For the next factorial moment
we obtain
$$\mathrm{E}[N(N-1)] =
\frac{1}{n^m} \times
m! [z^m]
\sum_{k=2}^n k (k-1) {n\choose k} (\exp(z)-1)^k
\\ = \frac{n(n-1)}{n^m} \times
m! [z^m]
\sum_{k=2}^n {n-2\choose k-2} (\exp(z)-1)^k
\\ = \frac{n(n-1)}{n^m} \times
m! [z^m] (\exp(z)-1)^2
\sum_{k=0}^{n-2} {n-2\choose k} (\exp(z)-1)^k
\\ = \frac{n(n-1)}{n^m} \times
m! [z^m] (\exp(z)-1)^2 \exp((n-2)z).$$
This is
$$\frac{n(n-1)}{n^m} (n^m - 2(n-1)^m + (n-2)^m)
\\ = n(n-1) \left(1-2\left(1-\frac{1}{n}\right)^m
+ \left(1-\frac{2}{n}\right)^m \right).$$
We get for the variance
$$\mathrm{Var}(N) = \mathrm{E}[N^2] - \mathrm{E}[N]^2
= \mathrm{E}[N(N-1)] + \mathrm{E}[N] - \mathrm{E}[N]^2$$
which yields
$$n^2 \left(1-2\left(1-\frac{1}{n}\right)^m
+ \left(1-\frac{2}{n}\right)^m \right)
\\ - n \left(1-2\left(1-\frac{1}{n}\right)^m
+ \left(1-\frac{2}{n}\right)^m \right)
+ n \left(1-\left(1-\frac{1}{n}\right)^m\right)
\\ - n^2 \left(1-2\left(1-\frac{1}{n}\right)^m
+ \left(1-\frac{1}{n}\right)^{2m} \right)
\\ = n^2 \left(\left(1-\frac{2}{n}\right)^m
- \left(1-\frac{1}{n}\right)^{2m} \right)
+ n \left(\left(1-\frac{1}{n}\right)^m
- \left(1-\frac{2}{n}\right)^m \right).$$
The absolute value of the two differences of terms that are geometric
in $m$ with positive common ratio less than one is bounded by the
maximum of these two, which vanishes in $m$ and hence the variance
goes to zero for $n$ fixed and $m$ going to infinity.
The queried standard deviation is then given by the root of the
variance.
Here is some Maple code to explore these numbers.
ENUM :=
proc(n, m)
option remember;
local ind, data, gf;
gf := 0;
for ind from n^m to 2*n^m - 1 do
data := convert(ind, base, n)[1..m];
gf := gf + v^nops(convert(data, `multiset`));
od;
gf/n^m;
end;
EN_PROB := (n, m) -> subs(v=1, ENUM(n, m));
EN_FM1 := (n, m) -> subs(v=1, diff(ENUM(n, m), v));
EN_FM2 := (n, m) -> subs(v=1, diff(ENUM(n, m), v$2));
FM1 := (n, m) -> n*(1-(1-1/n)^m);
FM2 := (n, m) -> n*(n-1)*(1 - 2*(1-1/n)^m + (1-2/n)^m);
EN_VAR := (n, m) -> - EN_FM1(n, m)^2
+ subs(v=1, diff(v*diff(ENUM(n, m), v), v));
VAR := (n, m) ->
n^2*((1-2/n)^m-(1-1/n)^(2*m)) + n*((1-1/n)^m-(1-2/n)^m);
Addendum. As pointed out by @JMoravitz we may need some
clarification of the reasoning by which the probabilities are
obtained. Note that $k$ distinct outcomes first of all require a
choice of these $k$ values from the $n$ possibilities, for a factor of
${n\choose k}.$ Now we need to match up each of these $k$ values
(think of them as listed sequentially from smallest to largest) with a
subset of the $m$ rolls which may not be empty (this is what prevents
double counting here). This yields the labeled combinatorial class (labeled means EGFs)
$$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}
\textsc{SEQ}_{=k}(\textsc{SET}_{\ge 1}(\mathcal{Z}))$$
or alternatively (compare to Stirling numbers of the second kind)
$$\textsc{SEQ}_{=k}(\textsc{SET}_{=1}(\mathcal{Z})
+ \textsc{SET}_{=2}(\mathcal{Z})
+ \textsc{SET}_{=3}(\mathcal{Z}) + \cdots).$$
We get the generating function
$$\left(z+\frac{z^2}{2!}+\frac{z^3}{3!}+\frac{z^4}{4!}+\cdots\right)^k
= (\exp(z)-1)^k$$
and may then continue as before.
Addendum II. Another combinatorial class we can use here is
$$\textsc{SEQ}_{=n}(\mathcal{E} +
\mathcal{U}\times\textsc{SET}_{\ge 1}(\mathcal{Z})).$$
This gives the EGF
$$G(z, u) = (1+u(\exp(z)-1))^n
= \sum_{k=0}^n {n\choose k} u^k (\exp(z)-1)^k.$$
On evaluating
$$\left. \frac{\partial}{\partial u} G(z, u) \right|_{u=1}$$
using the second form we get the same sum as before. We can also give
an alternate evaluation
$$\frac{1}{n^m} \times m! [z^m]
\left. \frac{\partial}{\partial u} G(z, u) \right|_{u=1}
= \frac{n}{n^m} \times m! [z^m] \exp((n-1)z) (\exp(z) - 1)
\\ = \frac{n}{n^m} \times m! [z^m] (\exp(nz) - \exp((n-1)z))
= \frac{n}{n^m} (n^m - (n-1)^m)
\\ = n \left(1-\left(1-\frac{1}{n}\right)^m\right).$$
For the next factorial moment we need another derivative which is
$$n(n-1)(1+u(\exp(z)-1))^{n-2} (\exp(z)-1)^2.$$
Set $u=1$ to get
$$n(n-1) \exp((n-2)z)(\exp(2z)-2\exp(z)+1).$$
Extracting coefficients now yields
$$\frac{n(n-1)}{n^m} \left(n^m - 2 (n-1)^m + (n-2)^m\right)$$
and the rest is as in the first version. Note that we get a faster
enumeration routine if we admit the classification from the
introduction. The result is midway between enumeration and the closed
form and is shown below.
with(combinat);
ENUM2 :=
proc(n, m)
option remember;
local gf, part, psize, mset;
gf := 0;
part := firstpart(m);
while type(part, `list`) do
psize := nops(part);
mset := convert(part, `multiset`);
gf := gf + binomial(n, psize)
*m!/mul(p!, p in part)
*psize!/mul(p[2]!, p in mset)*v^psize;
part := nextpart(part);
od;
gf/n^m;
end;
Best Answer
We approach this problem from a combinatorial perspective. The number of $n$-roll sequences with $k$ distinct values ($1\le k\le6$), out of $6^n$ sequences total, is $$D(n,k)=\binom6kk!\left\{n\atop k\right\}=\binom6k\sum_{j=0}^k(-1)^{k-j}\binom kjj^n$$ where $\left\{n\atop k\right\}$ is the Stirling number of the second kind and counts the number of ways to partition the rolls into homogeneous subsets, and $\binom6kk!$ is the number of ways to fill those subsets with dice rolls. Letting $n$ vary across the positive integers we get $$D(n,1)=6$$ $$D(n,2)=15\cdot2^n-30$$ $$D(n,3)=-60\cdot2^n+20\cdot3^n+60$$ $$D(n,4)=90\cdot2^n-60\cdot3^n+15\cdot4^n-60$$ $$D(n,5)=-60\cdot2^n+60\cdot3^n-30\cdot4^n+6\cdot5^n+30$$ $$D(n,6)=15\cdot2^n-20\cdot3^n+15\cdot4^n-6\cdot5^n+6^n-6$$ $\frac{D(n,k)}{6^n}$ then gives the probability an $n$-roll sequence will have $k$ distinct values. The expected value of $D$ for a given $n$ is then $$\mu_n=\sum_{k=1}^6k\cdot\frac{D(n,k)}{6^n}=6\left(1-\left(\frac56\right)^n\right)$$ and the standard deviation is $$\sigma_n=\sqrt{\sum_{k=1}^6\frac{D(n,k)}{6^n}(k-\mu_n)^2}=\sqrt{\frac{5\cdot144^n-6\cdot150^n+180^n}{6^{3n-1}}}$$ Note that $$\lim_{n\to\infty}\mu_n=6\text{ and }\lim_{n\to\infty}\sigma_n=0$$ which match our intuition, since for large $n$ almost all roll sequences should contain all six outcomes.
The SymPy code that generated these results can be found here.