The probability that there is at least one triangle can be expressed using inclusion/exclusion:
$$
\mathbb{P}\biggl(\bigcup_{i=1}^n A_i\biggr) =\sum_{k=1}^n (-1)^{k-1}\sum_{\scriptstyle I\subset\{1,\ldots,n\}\atop\scriptstyle|I|=k} \mathbb{P}\left(\bigcap_{i\in I} A_i\right)\;,
$$
where the $A_i$ are the events of individual triangles existing. I don't see a way to evaluate this in closed form, but it can be used to expand the desired probability in powers of $p$, and truncating the sum at some $k$ yields successively improved bounds (upper bounds for odd $k$ and lower bounds for even $k$). In the asymptotic case of fixed $\lambda=np$ as $n\to\infty$, the series can be summed exactly and yields a probability of $1-\mathrm e^{-\lambda^3/6}$.
The $k=1$ term is just the expectation value that Douglas gave in his now-deleted answer, $\binom n3p^3$, which is an upper bound for the desired probability by the first moment method (which can be regarded as a special case of the present approach).
The term for $k=2$ triangles yields contributions with $5$ and $6$ edges. With $5$ edges, there are $\binom n4$ ways to select four vertices and $6$ ways to select one of the $6$ edges between them that isn't used, for a contribution of $6\binom n4p^5$. With $6$ edges, we can either have $5$ different vertices, with $5$ options to choose the shared vertex and $\frac12\binom42=3$ options to choose pairs among the remaining $4$, for a contribution of $15\binom n5p^6$, or we can have $6$ different vertices, with $\frac12\binom63=10$ options to choose triangles among them, for a contribution of $10\binom n6p^6$. Thus the desired probability $q$ is bounded by
$$
\binom n3p^3-\left(6\binom n4p^5+15\binom n5p^6+10\binom n6p^6\right)\le q\le\binom n3p^3\;.
$$
Asymptotically, if you keep $\lambda=np$ constant as $n\to\infty$, this becomes
$$\frac{\lambda^3}6-\frac{\lambda^6}{72}\lesssim q\lesssim\frac{\lambda^3}6\;.$$
For $k$ edges, the only contribution with enough powers of $n$ to match the $k$ powers of $p$ is one where there is only one edge per vertex, namely where the $k$ edges form $k/3$ triangles among $k$ vertices. Thus, asymptotically, the desired probability is given by
$$
q\sim\sum_{j=1}^\infty(-1)^{j-1}\frac{\lambda^{3j}}{6^jj!}=1-\mathrm e^{-\lambda^3/6}\;.
$$
In the original form of your question, in which you had $p=1/n$ and thus $\lambda=1$, this would be $1-\mathrm e^{-1/6}\approx0.1535$.
If you want the full finite-size expansion in powers of $p$ up to $6$-th order, you need to include the contribution $\binom n4p^6$ from all $6$ edges formed by $4$ vertices to get
$$
q = \binom n3p^3-\left(6\binom n4p^5+15\binom n5p^6+10\binom n6p^6+\binom n4p^6\right)+O\left(p^7\right)\;.
$$
If there are $3 \binom n3 p^3$ triangles in expectation, and $3 \binom n3 p^2$ connected triples, the global clustering coefficient should approach their ratio (which is $p$).
Of course, naively taking their ratios doesn't work: $\mathbb E[\frac {X}Y]$ is not the same thing as $\frac{\mathbb E[X]}{\mathbb E[Y]}$. This is one of the main challenges in dealing with the expected value of a ratio. Instead, we'll show that both quantities are concentrated around their mean, and proceed that way.
Let $X$ denote the number of triangles in $\mathcal G(n,p)$. It's easy to see if we properly define triangles that $\mathbb E[X] = 3\binom n3 p^3$, which for consistency with connected triplets I want to define as $3\binom n3$ choices of a potential path $P_3$, and a $p^3$ chance that both edges of the path and the edge that makes it a triangle are present.
Moreover, the number of triangles is $3n$-Lipschitz in the edges of the graph (changing one edge changes the number of triangles by at most $3n$) so by McDiarmid's inequality
$$
\Pr[|X - \mathbb E[X]| \ge n^{2.5}] \le 2 \exp \left(-\frac{2n^5}{\binom n2 (3n)^2}\right) \le 2 e^{-4n/9}.
$$
If we let $Y$ be the number of connected triplets, then the expected number of them is $3\binom n3 p^2$: there are $3\binom n3$ potential copies of $P_3$ in the graph, and each is realized with a probability of $p^2$. This is actually $2n$-Lipschitz in the edges of the graph (each edge is part of at most $2n$ paths on $3$ vertices) but we'll round that up to $3n$-Lipschitz so that the same bound applies.
As a result, with probability $1 - 4 e^{-4n/9}$, both values are within $n^{2.5}$ of their expected value, and so the ratio $\frac{X}{Y}$ satisfies
$$
\frac{3\binom n3 p^3 - n^{2.5}}{3\binom n3 p^2 + n^{2.5}} \le \frac{X}{Y} \le \frac{3\binom n3 p^3 + n^{2.5}}{3\binom n3 p^2 - n^{2.5}}
$$
and therefore $\frac XY = p + O(n^{-0.5})$ in these cases. The remaining cases occur with probability $4e^{-4n/9}$, and the ratio is between $0$ and $1$ even then, so they can affect the expected value by at most $4e^{-4n/9}$ as well. Therefore $\mathbb E[\frac XY] = p + O(n^{-0.5})$, which approaches $p$ as $n \to \infty$.
Best Answer
Let $\cal{C}_{k+1}$ be defined
Then, for each graph $G$, the number of paths with $k$ edges, is precisely the number of $C_{k+1} \in \cal{C}_{k+1}$ such that, writing $C_{k+1}=(u_1,u_2,\ldots, u_{k+1})$ [keep in mind that $C_{k+1}$ is ordered], there is an edge between $u_i$ and $u_{i+1}$ for every $i=1,2,\ldots, k$. So now, let $G$ be a graph drawn according to $\cal{G}(n,\lambda/n)$. Then, for each $C_{k+1} \in \cal{C}_{k+1}$, let $A(C_{k+1})$ be the event
Then the number $N_k$ of paths with $k$ edges in $G$ satisfies $$N_k \ = \ \sum_{C_{k+1} \in \cal{C}_{k+1}} I_{A(C_{k+1})},$$ and so by linearity of expectation: $$\mathbb{E}[N_k] \ = \ \sum_{C_{k+1} \in \cal{C}_{k+1}} {\bf{P}}[A(C_{k+1})].$$
However, because $G$ was drawn according to $\cal{G}(n,\frac{\lambda}{n})$, the equation $${\bf{P}}[A(C_{k+1})] \ = \ \Big(\frac{\lambda}{n}\Big)^k$$ holds for each $C_{k+1} \in \cal{C}_{k+1}$. Furthermore, so does the equation $$|\cal{C}_{k+1}| = {n \choose {k+1}} \times (k+1)!$$ Putting these latter three equations together yields $$\mathbb{E}[N_k] \ = \ \Big(\frac{\lambda}{n}\Big)^k \times {n \choose {k+1}} \times (k+1)!$$ So the RHS of this is the expected number of paths with $k$ edges.