This is not a complete answer by any means, but is intended to get the ball rolling.

First of all, it need not be the case that $H^1(X_{fl},\mu_n) = 0.$ Rather, what follows
from the vanishing of $H^1(X_{fl},{\mathbb G}_m)$ is that
$H^1(X_{fl},\mu_n) = R^{\times}/(R^{\times})^n.$

(This is not always trivial; imagine
e.g. that $R$ is a non-algebraically closed field. You might have been thinking of the case
when $X$ is projective and smooth over an algebraically closed field, when the $H^0$-part of the exact sequence is itself exact, and so can be omitted from consideration. That is not the case here.)

Secondly, this doesn't hurt your arguments, because the same consideration of $H^0$-terms has to be made for the cohomology of $U$. Since $X$ is three dimensional and a complete intersection, restriction induces an isomorphism
$H^0(X,\mathcal O) \cong H^0(U,\mathcal O)$, and so also an isomorphism
$H^0(X,\mathcal O^{\times})\cong H^0(U,\mathcal O^{\times}),$ and so also isomorphisms
$H^0(X_{fl},\mu_n) \cong H^0(U_{fl},\mu_n)$ and
$H^0(X_{fl},{\mathbb G}_m)\cong H^0(U_{fl},{\mathbb G}_m).$ Thus in fact one finds that
the $n$-torsion in Pic$(U)$ is equal to the cokernel of the injection
$H^1(X_{fl},\mu_n) \hookrightarrow H^1(U_{fl},\mu_n).$ And as your analysis shows, this cokernel embeds into $H^2_{\{m\}}(X_{fl},\mu_n)$, with the cokernel of that embedding itself embedding into $H^2(X_{fl},\mu_n).$

So what can be said about this latter cohomology group?

Since $H^1(X_{fl},{\mathbb G}_m)$ vanishes, as you observed, one finds that $H^2(X_{fl}, \mu_n)$ coincides with the $n$-torsion in the cohomological Brauer group $H^2(X_{fl},{\mathbb G}_m).$ (Here I am using the fact that since ${\mathbb G}_m$ is smooth, flat and etale cohomology coincide, so $H^2(X_{fl},{\mathbb G}_m) = H^2(X_{et}, {\mathbb G}_m).$) So it seems that one wants to kill off the torsion in this Brauer group.

I don't see why this need be true, but what one actually needs is that $H^2(X_{fl},\mu_n)
\rightarrow H^2(U_{fl},\mu_n)$ is injective. Since $H^2(X_{fl},\mu_n)$ embeds into
$H^2(X_{fl},{\mathbb G}_m)$, it would be enough to show that the restriction
$H^2(X_{fl},{\mathbb G}_m) \to H^2(U_{fl},{\mathbb G}_m)$ induces an injection on torsion. Might this be some kind of purity result on Brauer groups of the kind Gabber discusses in his abstract? It would be related to a vanishing of (torsion in) $H^2_{\{m\}}(X_{fl}, {\mathbb G}_m)$. Somewhere (maybe here?) one presumably has to make use of the dimension and lci hypotheses.

P.S. You may well just want to email Gabber to ask him about this. If you do, and you get an answer, please share it!

EDIT: This is an excerpt from the email referred to in Hai Long's comment below:

To learn about these kinds of arguments, my advice is to
do just what you are doing. One works with the exact sequence
linking $\mu_n$ and ${\mathbb G}_m$, as you did.

Number theorists (at least of a certain stripe) have some advantages
with this, because the case $X =$ Spec $K$ ($K$ a field) comes up a lot
under the name of Kummer theory, and also Mazur in one of his
famous papers uses a lot of flat cohomology. But in the end, the
formalism is just the one you used in your question.

Then, typically, one has to inject something additional that
is less formal. My suggestion would be to look at de Jong's proof
of Gabber's result showing $Br'(X) = Br(X)$ discussed in his abstract.
(There is a write-up on de Jong's web-page.)

Reading the proof of a result like this might give some insight
into how to work with Brauer groups in a less formal way.