Your conjectures are correct.  So was the "someone else at MSRI [who]
muttered something about norm forms" (mentioned in earlier edits
of the question), except for the part about laughing at you.

As you in effect note, $f(a,b,c)$ is the norm $N_{K/{\bf Q}}(a+bx+cx^2)$,
where $x$ is one of the roots of $x^3-x^2-x-1 = 0$
and $K$ is the cubic number field ${\bf Q}(x)$.
This field has discriminant $-44$,
and ${\bf Z}[x]$ is the full ring of integers $O_K$
(equivalently, the field discriminant of $K/{\bf Q}$
equals the polynomial discriminant of $x^3-x^2-x-1$;
to check this in **gp**, compute

    poldisc(x^3-x^2-x-1)
    nfdisc(x^3-x^2-x-1)

and observe that both return $-44$).  Now for (A),
you already know that $x^3-x^2-x-1$ has at least one root modulo
any prime $q$ unless $q$ is represented by the nonprincipal quadratic
form $3u^2+2uv+4v^2$ of discriminant $-44$.  (For other $q$:
there's a triple root for $q=2$, a double and a simple root for $q=11$,
three distinct roots for $q=u^2+11v^2$, and one simple root for odd
$q$ not congruent to a square $\bmod 11$.)  Equivalently, $K$ has
an ideal of norm $q$ unless $q = 3u^2+2uv+4v^2$.  But $O_K$ is
a principal ideal domain, so once there's an ideal of norm $q$
then it has a generator $a+bx+cx^2 \in O_K$, and then $q=f(a,b,c)$
(or $q=f(-a,-b,-c)$ if we chose $a+bx+cx^2$ of norm $-q$).  The discriminant of $K$
is small enough that one can check unique factorization by hand using
the Minkowski bound; nowadays this exercise can also be done routinely
on the computer, e.g. in **gp**

    K = bnfinit(x^3-x^2-x-1); K.cyc

(This functionality happens to be one of the "Usage examples" in the current
<a href="http://en.wikipedia.org/wiki/PARI/GP#Usage_examples">Wikipedia
page on **gp**</a>.)

(B) Translating the factorization of $x^3-x^2-x-1 \bmod q$
into the factorization of the ideal $(q)$ in $O_K$, we see that
if $q = 3u^2+2uv+4v^2$ then $(q)$ remains prime in $O_K$,
and thus that $q \mid N_{K/{\bf Q}}(a+bx+cx^2)$ **iff** $q \mid a+bx+cx^2$.
For $q=2$ the ideal $(q)$ is the cube of $(1+x)$, so $8 \mid f(a,b,c)$
**iff** $a,b,c$ are all even.  Any power of a prime $q$ other than
those of the form $3u^2+2uv+4v^2$ can be represented primitively by $f$,
even $q=11$ (for which $(q)$ factors as $(2+x)(3-2x)^2$).  If we do not
care about primitivity then we can also represent all powers of $2$,
and all powers of $q^3$ for $q = 3u^2+2uv+4v^2$.

By multiplicativity this also proves the final conjecture:
the nonzero $n \in {\bf Z}$ that are represented by $f$
are precisely those whose $q$-valuation is a multiple of $3$
for all primes $q = 3u^2+2uv+4v^2$.