TQFT is most interesting when we're allowed to vary the topology of the manifold, but in ordinary QFT we normally consider the QFT on a fixed spacetime manifold. For a fair comparison, I'll consider a TQFT on a fixed manifold (without metric, of course).
Disclaimer: I'm a relative newcomer to the subjects of TQFT and higher-form symmetries. This answer reflects my current understanding, but I might be missing something.$^\dagger$
$^\dagger$ Edit: Turns out that I was missing something, but maybe it doesn't completely invalidate the conclusion. See the comment by Ryan Thorngren for details.
The answer is yes
Let $S$ be the set of observables of this TQFT, represented as operators on a Hilbert space ${\cal H}$. To define spontaneous symmetry breaking (SSB), first we need to define symmetry. Suppose we define a "symmetry" to be any unitary transformation $U$ that preserves the set $S$ but that has a non-trivial effect on at least one observable in $S$. Then we can define SSB to be the condition that at least one ground state is not invariant under $U$.
With this definition of symmetry, every non-trivial symmetry of a TQFT is spontaneously broken, simply because every state in ${\cal H}$ is a ground state, so if any state in ${\cal H}$ is not invariant under $U$, we could call it SSB. If $U$ is a non-trivial symmetry (not the identity operator), then ${\cal H}$ must have at least one state that is not invariant under $U$.
Does this definition of SSB agree with the one we normally use in ordinary QFT? I think it does, because of the phrase "at least one ground state." Even if the symmetry group in question is $\mathbb{Z}_2$, so that we can take the direct sum of the two SSB Hilbert spaces and construct a ground state that is invariant under the symmetry (disregarding the cluster property), the theory still admits at least one ground state (in some representation) that is not invariant under $U$. So I the definition I described above is consistent with the usual one.
... or is the answer no (for conventional symmetries)?
On the other hand, page 3 in Ryan Thorngren's paper https://arxiv.org/abs/2001.11938 says
Nontrivial $d-1$-form symmetries of TQFTs in $d+1$ dimensions are always spontaneously broken... [and] $0$-form symmetries of TQFTs are always unbroken...
A $0$-form symmetry is a symmetry in the conventional sense. On page 2, the cited paper defines SSB in terms of long range order. For a $0$-form symmetry, the definition of long-range order relies on local observables, and since local observables don't exist in a TQFT, we immediately conclude that $0$-form symmetries are never spontaneously broken in TQFT, as stated in the excerpt.
Reconciling the two conclusions
The two different definitions of SSB shown above might seem to lead to opposite conclusions: one says that non-trivial symmetries in TQFT are always spontaneously broken, and the other says that conventional ($0$-form) symmetries in TQFT are never spontaneously broken. And yet, if I'm not mistaken, both definitions agree with the one we would normally use for conventional ($0$-form) symmetries in ordinary QFT.
How is this possible? If both definitions agree with the one we normally use in ordinary QFT, then how can they give different answers in a TQFT? After all, we can get a TQFT by taking the extreme low-energy limit of an ordinary gapped QFT. What's going on here?
I think$^\dagger$ this is resolved by recognizing that a "symmetry" according to the first definition is never a $0$-form symmetry. It can't be, because a non-trivial $0$-form symmetry must (by definition) have a non-trivial effect on local observables (observables localized in a contractible region of the spacetime), but a TQFT doesn't have any local observables for the symmetry to affect. The first definition implicitly catches all of the theory's symmetries, including $k$-form symmetries for $k\geq 1$, so it catches the fact that non-trivial $k$-form symmetries in TQFT can be spontaneously broken. When we take the extreme low-energy limit of an ordinary gapped QFT, we lose all of the local observables, so whatever $0$-form symmetries the theory had become trivial, whether or not they were spontaneously broken before the limit.
Altogether, the answer is yes: a TQFT can have SSB, if we consider $k$-form symmetries for $k\geq 1$. If we only consider conventional ($0$-form) symmetries, then the answer is no: a TQFT can't have SSB for a $0$-form symmetry simply because it can't have any non-trivial $0$-form symmetries (broken or not).
$^\dagger$ Edit: The reasoning in these last two paragraphs is incorrect, as clarified by Ryan Thorngren's comment. TQFTs can have non-trivial $0$-form symmetries. That makes today a good day — I learned something new!
