There are $k(G)|G|$ commuting pairs $(y,z)$ of elements of $G.$ How many commuting triples $(x,y,z)$ of elements of $G$ are there? If we fix the first component $x$ of the triple, note that $x$ permutes (by conjugation) the commuting pairs of elements of $G,$ and the only such pairs it fixes are the commuting pairs which already have both components in $C_{G}(x).$ Hence $x$ fixes $k(C_{G}(x))|C_{G}(x)|$ commuting pairs, and the total number of commuting triples in $G \times G \times G$ is $\sum_{x \in G} k(C_{G}(x))|C_{G}(x)|$. If $G$ has $k$ conjugacy classes, with representatives $\{ x_{i} : 1 \leq i \leq k \}$, this may be rewritten as $|G| \sum_{i=1}^{k} k(C_{G}(x_{i})).$ The probability you require, assuming a uniform distribution, is $\frac{1}{|G|^{2}}\left( \sum_{i=1}^{k} k(C_{G}(x_{i})) \right)$. The discussion (and Burnside's counting lemma) makes it clear that this is (number of orbits of $G$ by conjugation on commuting pairs of elements of $G$), divided by $|G|^{2}$.

There are $k(G)|G|$ commuting pairs $(y,z)$ of elements of $G.$ How many commuting triples $(x,y,z)$ of elements of $G$ are there? If we fix the first component $x$ of the triple, note that $x$ permutes (by conjugation) the commuting pairs of elements of $G,$ and the only such pairs it fixes are the commuting pairs which already have both components in $C_{G}(x).$ Hence $x$ fixes $k(C_{G}(x))|C_{G}(x)|$ commuting pairs, and the total number of commuting triples in $G \times G \times G$ is $\sum_{x \in G} k(C_{G}(x))|C_{G}(x)|$. If $G$ has $k$ conjugacy classes, with representatives $\{ x_{i} : 1 \leq i \leq k \}$, this may be rewritten as $|G| \sum_{i=1}^{k} k(C_{G}(x_{i})).$ The probability you require, assuming a uniform distribution, is $\frac{1}{|G|^{2}}\left( \sum_{i=1}^{k} k(C_{G}(x_{i})) \right)$. The discussion (and Burnside's counting lemma) makes it clear that this is (number of orbits of $G$ by conjugation on commuting pairs of elements of $G$), divided by $|G|^{2}$.

Later edit: Since I noticed the "(or more)" in the question title, the pattern is now clear: let $c_{n}(G)$ denote the number of commuting (ordered) $n$-tuples of elements of $G$. Then we have $c_{n+1}(G) = \sum_{x \in G} c_{n}(C_{G}(x))$.

There are $k(G)|G|$ commuting pairs $(y,z)$ of elements of $G.$ How many commuting triples $(x,y,z)$ of elements of $G$ are there? If we fix the first component $x$ of the triple, note that $x$ permutes the commuting pairs of elements of $G,$ and the only such pairs it fixes are the commuting pairs which already have both components in $C_{G}(x).$ Hence $x$ fixes $k(C_{G}(x))|C_{G}(x)|$ commuting pairs, and the total number of commuting triples in $G \times G \times G$ is $\sum_{x \in G} k(C_{G}(x))|C_{G}(x)|$. If $G$ has $k$ conjugacy classes, with representatives $\{ x_{i} : 1 \leq i \leq k \}$, this may be rewritten as $|G| \sum_{i=1}^{k} k(C_{G}(x_{i})).$ The probability you require, assuming a uniform distribution, is $\frac{1}{|G|^{2}}\left( \sum_{i=1}^{k} k(C_{G}(x_{i})) \right)$.

There are $k(G)|G|$ commuting pairs $(y,z)$ of elements of $G.$ How many commuting triples $(x,y,z)$ of elements of $G$ are there? If we fix the first component $x$ of the triple, note that $x$ permutes (by conjugation) the commuting pairs of elements of $G,$ and the only such pairs it fixes are the commuting pairs which already have both components in $C_{G}(x).$ Hence $x$ fixes $k(C_{G}(x))|C_{G}(x)|$ commuting pairs, and the total number of commuting triples in $G \times G \times G$ is $\sum_{x \in G} k(C_{G}(x))|C_{G}(x)|$. If $G$ has $k$ conjugacy classes, with representatives $\{ x_{i} : 1 \leq i \leq k \}$, this may be rewritten as $|G| \sum_{i=1}^{k} k(C_{G}(x_{i})).$ The probability you require, assuming a uniform distribution, is $\frac{1}{|G|^{2}}\left( \sum_{i=1}^{k} k(C_{G}(x_{i})) \right)$. The discussion (and Burnside's counting lemma) makes it clear that this is (number of orbits of $G$ by conjugation on commuting pairs of elements of $G$), divided by $|G|^{2}$.