Taking an algebraically closed field $K$ with $char(K)=0$ and $|K|\geq |\mathbb{C}|$, we have that $\bar{\mathbb{Q}}\subset K$. Let $T$ be a transcendence basis of $K\setminus \bar{\mathbb{Q}}$. Writing $|T|=\alpha$ and $[K:\bar{\mathbb{Q}}(T)]=n$, I wonder what is $|K|$ in terms of $\alpha$ and $n$. Does $|K|$ determine $\alpha$ and $n$ uniquely?
Actually, I want to prove that any two algebraically closed fields with $0$ characteristic and same cardinality greater or equal then $|\mathbb{C}|$ are isomorphic and I already know that $\alpha$ and $n$ determine $K$ up to isomorphisms. The question is the missing step.
Algebraic extensions never change the cardinality of an infinite field. Indeed, if $E$ is an infinite field and $F$ is an algebraic extension, every $x\in F$ is a root of some polynomial over $E$. There are only $|E|$ polynomials over $E$ and each polynomial has only finitely many roots, so $F$ cannot have more than $|E|$ elements.
So, in your setup, we have $|K|=|\bar{\mathbb{Q}}(T)|$. Moreover, $|\bar{\mathbb{Q}}(T)|=\aleph_0+|T|$, since every element of $\bar{\mathbb{Q}}(T)$ is a rational function in elements of $T$ with coefficients in $\bar{\mathbb{Q}}$ and $\bar{\mathbb{Q}}$ is countable. In your case, $T$ must be uncountable since $|K|\geq|\mathbb{C}|$, so we get $|K|=|\bar{\mathbb{Q}}(T)|=|T|$.
More generally, the same argument shows that if $K$ is any uncountable field, then its cardinality is equal to its transcendence degree over the prime field.
What you call $n$ must also always be equal to $|T|$. For instance, for any $t\in T$, there is a square root $\sqrt{t}$ in $K$, and these elements are linearly independent over $\bar{\mathbb{Q}}(T)$. So $n\geq |T|$, but also $n\leq |K|=|T|$ so $n=|T|$