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 Wikipedia page on gp.)

(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 $q = 3u^2+2uv+4v^2$.

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 Wikipedia page on gp.)

[EDIT In fact this $K$ happens to be one of the handful of number fields whose Minkowski bound is so tight that nothing needs to be checked! The discriminant $\Delta_K = -44$ is small enough in absolute value that the bound $$ \frac4\pi \frac{3!}{3^3} \left|\Delta_K\right|^{1/2} = 1.8768\ldots $$ is less than $2$, which means every ideal $I$ has a nonzero element of norm $\pm \left|I\right|$ and is thus automatically principal. TIDE]

(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 $q = 3u^2+2uv+4v^2$.