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)\;. $