$\newcommand{\cl}{\operatorname{cl}}$
Let $R \subset S \times S$ be a binary relation, the smallest
i) reflexive relation containing it is $$ \cl_\mathrm{ref} = R \cup \{ (x,x) : x \in S \} $$ ii) symmetric relation $$ \cl_\mathrm{sym} = R \cup \{ (y,x) : (x,y) \in R \} $$ ii) transitive relation $$ \cl_\mathrm{trans}(R) = R \cup \{ (x_1, x_n) : (x_1, x_2), \ldots, (x_{n-1}, x_n) \in R \} = \bigcup_{i = 1}^\infty R^i. $$ And the smallest equivalence relation containing it is $$ \cl_\mathrm{ref}(R) \cup \cl_\mathrm{sym}(R) \cup \cl_\mathrm{trans}(R) $$
But how looks the smallest congruence relation on a set $R$ with respect to some $n$-ary operation $f$? Is there a similar "constructive" way of defining it?
EDIT: Formula for transitive closure is wrong, see comments. Correct version $$ \cl_\mathrm{trans}(\cl_\mathrm{sym}(\cl_\mathrm{ref}(R))). $$
Let $\mathfrak{S}=(S,F)$ be an algebra where $S$ is the universe and $F$ is the collection of fundamental operations on $S$. For an arbitrary $R \subseteq S \times S$, define $R\circ R$ to be $\{(x,z): (x,y),(y,z)\in R\}$.
Define $R_0=R\cup \{(x,x):x\in S\} \cup \{(y,x):(x,y)\in R\} = \mathrm{cl}_{sym}(\mathrm{cl}_{ref}(R))$.
Then define $R_{n+1} = (R_n \circ R_n) \cup \{\big(f(x_1,\ldots,x_n),f(y_1,\ldots,y_n)\big):f\in F \text{ and } (x_1,y_1),\ldots, (x_n,y_n) \in R_n\}$.
Then $\displaystyle\bigcup_{i=0}^\infty R_i$ is the smallest congruence relation containing $R$.