As an immediate consequence of Proposition 6.5.2 of Thurston's notes, we have that, if $M$ is a compact 3-manifold with toric boundary and $\tilde M$ is obtained from $M$ via Dehn filling, then for the 'variant' $\|\cdot\|_0$ of the Gromov norm it holds that $\|[\tilde M,\partial\tilde M]\|_0\leq \|[M,\partial M]\|_0$. Under which assumptions is it also true that $\|[\tilde M,\partial\tilde M]\|\leq \|[M,\partial M]\|$, $\|\cdot\|$, being the 'proper' Gromov norm? This holds, for instance, if $M$ is hyperbolic (Lemma 6.5.4), but what if we drop this hypothesis?

As mentioned in another answer, one may deduce the desired result from the following facts:

(Gromov's Equivalence Theorem): If every component of the boundary of $M$ has amenable fundamental group, then for every $\varepsilon>0$ there exists a fundamental cycle $z$ for $M$ such that $\|z\|\leq \|M,\partial \|+\varepsilon$ and $\|\partial z\|\leq \varepsilon$.

(Matsumoto-Morita): If $T$ is a boundary component of $M$ and the fundamental group of $T$ is amenable, then any two fundamental cycles $z,z'$ for $T$ differ by a boundary $c$ such that $\|c\|\leq K\cdot (\|z\|+\|z'\|)$.

Putting together (1) and (2) one easily gets that the simplicial volume is subadditive with respect to gluings along boundaty components having amenable fundamental groups.

Gromov's proof of the Equivalence Theorem is based on the quite technical theory of multicomplexes, which is studied and developed in the preprint by Kuessner cited above. A different proof (using Ivanov's and Monod's approach to bounded cohomology) is given in our recent preprint

http://arxiv.org/abs/1305.2612

(joint work with Bucher, Burger, Iozzi, Pagliantini, Pozzetti).

This is certainly true, and follows by consideration of the decomposition of a manifold (let's say irreducible, to simplify the discussion) M along incompressible tori into hyperbolic and Seifert fibered pieces. The Gromov norm is the sum of the norms of the hyperbolic pieces. Now, if you fill a torus that is a boundary component of a hyperbolic piece, the norm decreases (strictly, in fact). If you fill a torus that is part of a Seifert fibered piece, then if the component that you filled remains Seifert fibered with its boundary components incompressible, the norm of the manifold is unchanged. Otherwise, you may have turned your Seifert fibered manifold into a solid torus, in which case you repeat the argument; in this setting the Gromov norm may stay the same or strictly decrease.

An example of the latter setting where the Gromov norm might strictly decrease is the following. Consider a hyperbolic manifold N with a single cusp, glued to C = complement of a torus knot in a solid torus, along the boundary of the solid torus. If you fill the torus knot complement using the meridian slope, then C becomes a solid torus, and the resulting manifold is then a filling of N. Thus the Gromov norm would strictly decrease.

In fact, for manifolds with torus boundary you have $$\parallel M,\partial M\parallel_0=\parallel M,\partial M\parallel$$ so that the two inequalities are plainly equivalent.

(More generally, this equality holds whenever all components of $\partial M$ have amenable fundamental group.)

There is also an explicit argument which shows $$\parallel M_1\cup_{A}M_2\parallel \le \parallel M_1,\partial M_1\parallel + \parallel M_2,\partial M_2\parallel$$ whenever $A=\partial M_1=\partial M_2$ has amenable fundamental group, see Lemma 5 in http://newton.kias.re.kr/~kuessner/preprints/bc.pdf

It applies to Dehn surgery with $M_2$ the solid torus, whose simplicial volume is zero.

The proof follows essentially from a result of Matsumoto-Morita (Proc. AMS 94, 539-544, 1985). They proved that (assuming amenable fundamental group) there exists a constant K such that for each boundary $z\in B_2(\partial M)$ one has a chain $c\in C_3(\partial M)$ with $\partial c=z$ and $\parallel c\parallel\le K\parallel z\parallel$. One can apply this to obtain fundamental cycles for $M_1$ and $M_2$ which have suitable norms and which fit together at the common boundary.