Let $k$ be a field, and $A$ a $k$-domain, so that the fraction field of $A$ has transcendence degree $n$ over $k$.

If $A$ is finitely-generated over $k$, then $A$ has Krull dimension $n$ (Theorem A in Eisenbud).

However, if $A$ is infinitely-generated, then it is possible for the dimension of $A$ to be less than the transcendence degree of its fraction field.  Take, for example, rational functions in one variable $A=k(x)$.  Dimension 0, transcendence degree 1.

Is it always true that the dimension of $A$ is less than or equal to $n$?

It looks to me like the answer is yes.

Fix any strictly increasing chain of primes $P_0 \subsetneq P_1 \subsetneq \cdots \subsetneq P_m$ in $A$ of length $m$; we'll prove that $m \leq n$. Choose elements $x_i \in P_i \setminus P_{i-1}$ for $i = 1,\dots,m$. 

Let $B \subseteq A$ be the $k$-subalgebra generated by $\{x_1,\dots,x_m\}$.  The primes $Q_i = P_i \cap B$ form a strictly increasing chain of length $m$ in $B$, because $x_i \in Q_i \setminus Q_{i-1}$.

The fraction field $Q(B)$ of $B$ lives inside of $Q(A)$ and therefore has $k$-transcendence degree at most $n$.  Because $B$ is finitely generated, this means that it has Krull dimension at most $n$. It now follows that $m \leq n$.
