All the answers so far have focused on 3 dimensions, but the answer is much more striking in 4 dimensions. Freedman's theorem tells you that classical homology invariants give you complete information about topological, simply-connected 4-manifolds. These classical invariants cannot, however, distinguish between distinct smooth structures on the same topological 4-manifold, and essentially our only technique for distinguishing smooth 4-manifolds is Donaldson's invariant or the Seiberg-Witten invariant or their relatives. These do not quite form a TQFT, but are related to TQFTs.

Edit: On request, a little about how the 4-manifold invariants are related to a TQFT. This is all nicely explained in the beginning of Kronheimer and Mrowka's book Monopoles and 3-manifolds.

There are actually three different theories, denoted $\widehat{\mathit{HM}}$ ("HM-from"), $\check{\mathit{HM}}$ ("HM-to", unfortunately typeset badly here), and $\overline{\mathit{HM}}$. All are close to satisfying axioms for a TQFT assigning a vector space to a 3-manifold and maps to a 4-manifold, at least for connected manifolds. (The vector spaces are infinite dimensional, but finite in each graded piece.) Unfortunately, however you slice it, in each case the invariant associated to a closed 4-manifold in the usual TQFT way (when defined) is zero.

Instead, you use the fact that there is an exact triangle \[ \cdots \longrightarrow \widehat{\mathit{HM}} \longrightarrow \overline{\mathit{HM}} \longrightarrow \check{\mathit{HM}}\longrightarrow \cdots \] (with right mapping to left), and the map $\overline{\mathit{HM}}(W)$ is $0$ for $b_2^+(W) \ge 1$.

If you have a 4-manifold $W$ with $b_2^+(W) \ge 2$, you factor it as two cobordisms $W = W_1 \cup_Y W_2$ for some 3-manifold $Y$, with $b_2^+(W_i) \ge 1$. Then the properties above let you map from $\check{\mathit{HM}}(S^3)$, to $\check{\mathit{HM}}(Y)$, backwards in the exact triangle to $\widehat{\mathit{HM}}(Y)$, and then forwards to $\widehat{\mathit{HM}}(S^3)$. The resulting map (from $\check{\mathit{HM}}(S^3)$ to $\widehat{\mathit{HM}}(S^3)$) gives the interesting Seiberg-Witten invariants of $W$.

All the answers so far have focused on 3 dimensions, but the answer is much more striking in 4 dimensions. Freedman's theorem tells you that classical homology invariants give you complete information about topological, simply-connected 4-manifolds. These classical invariants cannot, however, distinguish between distinct smooth structures on the same topological 4-manifold, and essentially our only technique for distinguishing smooth 4-manifolds is Donaldson's invariant or the Seiberg-Witten invariant or their relatives. These do not quite form a TQFT, but are related to TQFTs.

Edit: On request, a little about how the 4-manifold invariants are related to a TQFT. This is all nicely explained in the beginning of Kronheimer and Mrowka's book Monopoles and 3-manifolds.

There are actually three different theories, denoted $\widehat{\mathit{HM}}$ ("HM-from"), $\check{\mathit{HM}}$ ("HM-to", unfortunately typeset badly here), and $\overline{\mathit{HM}}$. All are close to satisfying axioms for a TQFT assigning a vector space to a 3-manifold and maps to a 4-manifold, at least for connected manifolds. (The vector spaces are infinite dimensional, but finite in each graded piece.) Unfortunately, however you slice it, in each case the invariant associated to a closed 4-manifold in the usual TQFT way (when defined) is zero.

Instead, you use the fact that there is an exact triangle $$ \cdots \longrightarrow \widehat{\mathit{HM}} \longrightarrow \overline{\mathit{HM}} \longrightarrow \check{\mathit{HM}}\longrightarrow \cdots $$ (with right mapping to left), and the map $\overline{\mathit{HM}}(W)$ is $0$ for $b_2^+(W) \ge 1$.

If you have a 4-manifold $W$ with $b_2^+(W) \ge 2$, you factor it as two cobordisms $W = W_1 \cup_Y W_2$ for some 3-manifold $Y$, with $b_2^+(W_i) \ge 1$. Then the properties above let you map from $\check{\mathit{HM}}(S^3)$, to $\check{\mathit{HM}}(Y)$, backwards in the exact triangle to $\widehat{\mathit{HM}}(Y)$, and then forwards to $\widehat{\mathit{HM}}(S^3)$. The resulting map (from $\check{\mathit{HM}}(S^3)$ to $\widehat{\mathit{HM}}(S^3)$) gives the interesting Seiberg-Witten invariants of $W$.

All the answers so far have focused on 3 dimensions, but the answer is much more striking in 4 dimensions. Freedman's theorem tells you that classical homology invariants give you complete information about topological, simply-connected 4-manifolds. These classical invariants cannot, however, distinguish between distinct smooth structures on the same topological 4-manifold, and essentially our only technique for distinguishing smooth 4-manifolds is Donaldson's invariant or the Seiberg-Witten invariant or their relatives. These do not quite form a TQFT, but are related to TQFTs.

All the answers so far have focused on 3 dimensions, but the answer is much more striking in 4 dimensions. Freedman's theorem tells you that classical homology invariants give you complete information about topological, simply-connected 4-manifolds. These classical invariants cannot, however, distinguish between distinct smooth structures on the same topological 4-manifold, and essentially our only technique for distinguishing smooth 4-manifolds is Donaldson's invariant or the Seiberg-Witten invariant or their relatives. These do not quite form a TQFT, but are related to TQFTs.

Edit: On request, a little about how the 4-manifold invariants are related to a TQFT. This is all nicely explained in the beginning of Kronheimer and Mrowka's book Monopoles and 3-manifolds.

There are actually three different theories, denoted $\widehat{\mathit{HM}}$ ("HM-from"), $\check{\mathit{HM}}$ ("HM-to", unfortunately typeset badly here), and $\overline{\mathit{HM}}$. All are close to satisfying axioms for a TQFT assigning a vector space to a 3-manifold and maps to a 4-manifold, at least for connected manifolds. (The vector spaces are infinite dimensional, but finite in each graded piece.) Unfortunately, however you slice it, in each case the invariant associated to a closed 4-manifold in the usual TQFT way (when defined) is zero.

Instead, you use the fact that there is an exact triangle \[ \cdots \longrightarrow \widehat{\mathit{HM}} \longrightarrow \overline{\mathit{HM}} \longrightarrow \check{\mathit{HM}}\longrightarrow \cdots \] (with right mapping to left), and the map $\overline{\mathit{HM}}(W)$ is $0$ for $b_2^+(W) \ge 1$.

If you have a 4-manifold $W$ with $b_2^+(W) \ge 2$, you factor it as two cobordisms $W = W_1 \cup_Y W_2$ for some 3-manifold $Y$, with $b_2^+(W_i) \ge 1$. Then the properties above let you map from $\check{\mathit{HM}}(S^3)$, to $\check{\mathit{HM}}(Y)$, backwards in the exact triangle to $\widehat{\mathit{HM}}(Y)$, and then forwards to $\widehat{\mathit{HM}}(S^3)$. The resulting map (from $\check{\mathit{HM}}(S^3)$ to $\widehat{\mathit{HM}}(S^3)$) gives the interesting Seiberg-Witten invariants of $W$.