As you mention, the Akra–Bazzi theorem shows that the solution to the recurrence $T(n,p)$ is $O(n\log n)$ for all $p \in (0,1)$. However, this does not reveal the nature of the dependence on $p$. To determine the latter, we can use a recursion tree approach.
At the root of the recursion tree is the interval $\{1,\ldots n\}$. Its two children are the intervals $\{1,\ldots,pn\}$ and $\{pn+1,\ldots,n\}$, whose total length is again $n$. Each of these nodes has two children (assuming $n$ is large enough), and so on. For simplicity we ignore rounding errors, that is, we assume that $pn$ is an integer; this is just a technicality, and I wouldn't worry about it. We stop the process whenever a node has length at most $1$. The complexity of the algorithm is proportional to the total length of intervals in the tree. When $p \neq 1/2$, the leaves (nodes at which we stop the process) have different depth, and that makes it more difficult to determine the overall complexity.
We can obtain a simple upper bound by noting that the tree has at most $\log_{1-p} (1/n)$ levels: each node is at least a factor of $1-p$ smaller than its parent. Just like in the analysis for $p = 1/2$, the total length of intervals at any level is at most $n$, and we obtain an upper bound of $O(n\log_{1-p} (1/n))$ on the running time. Since $\log_{1-p} (1/n) = \log n/\log (1-p)^{-1}$ and $\log (1-p)^{-1} = -\log (1-p) = p \pm O(p^2)$ for small $p$, we can write this as $O(n\log n/p)$.
Here is a more accurate calculation. Consider level $t$. Suppose we don't stop the process upon reaching a small interval. We can generate a random vertex by taking $t$ steps, in each of which we go left (say) with probability $p$ and right (say) with probability $1-p$. Each time we take a left step the log of the length of the interval decreases by $-\log p$, and each time we take a right step it decreases by $-\log (1-p)$. A vertex is in the actual tree of the log of the length decreased by at most $\log n$. The total weight of intervals on level $t$ of the tree is exactly the probability that a vertex generated according to this process correspond to a decrease of at most $\log n$. That is, if $D$ is the distribution which is equal to $-\log p$ with probability $p$ and to $-\log(1-p)$ with probability $1-p$, and $X_1,\ldots,X_t \sim D$ are independent, then the total weight of level $t$ is $\Pr[X_1+\cdots+X_t \leq \log n]$. For super-constant $t$, the random variable $X_1+\cdots+X_t$ is roughly normally distributed with mean $[-p\log p-(1-p)\log(1-p)]t$ and variance linear in $t$, so for $t$ satisfying $[-p\log p-(1-p)\log(1-p)]t \leq (\log n)/2$, say, the probability will be very close to $1$, while for $t$ satisfying $[-p\log p-(1-p)\log(1-p)]t \geq 2\log n$, say, it will be very close to zero. Defining $h(p) = -p\log p-(1-p)\log(1-p)$ (known as the binary entropy function), we conclude that the running time is $\Theta(n\log n/h(p))$ (uniform in $p$, as $n\to\infty$). As $p\to 0$ we have $h(p) \approx -p\log p$, and so our earlier estimate was not tight.
Another way of looking at the same analysis is by having an infinite sequence of independent random variables $X_1,X_2,\ldots$ as before, and defining a stopping time $T$ to be the first time $t$ such that $X_1 + \cdots + X_t \geq \log n$. The running time is then proportional to $n\mathbb{E}[T]$. The elementary renewal theorem then states that $\lim_{n\to\infty} \mathbb{E}[T]/\log n = 1/\mathbb{E}[D] = 1/h(p)$, implying that the total size of intervals equals $(1+o(1))n\log n/h(p)$. More accurately, for every constant $p$ the total size of intervals is $(1+\alpha_p(n))n\log n/h(p)$, where $\alpha_p(n) = o(n)$. The convergence in the elementary renewal theorem is exponential in the time parameter — $\log n$ in our case — so should be polynomial in $n$, that is, $\alpha_p(n) = O(n^{-C_p})$. The convergence is also probably uniform for $p \in (\delta,1-\delta)$ for any $\delta > 0$.
Summarizing, the total length of intervals in the recursion tree, which is proportional to the running time, is of the following form for every $p$: $$ T(n,p) = (1+o(1)) \frac{n\log n}{h(p)}, $$ where $\log n$ and $h(p) = -p\log p-(1-p)\log(1-p)$ are taken to the same base, and the $o(1)$ is a function depending on $p$ and tending to $0$ with $n$.
Moreover, it is probably true that for any $\delta > 0$ and any $p \in (\delta,1-\delta)$ it is true that the total length of intervals is of the form $$T(n,p) = (1+O(n^{-C_\delta})) \frac{n\log n}{h(p)}, $$ where $C_\delta > 0$ and the hidden big O constant depend only on $\delta$. In particular, it should be the case that for all constant $p_1,p_2$,
$$\lim_{n\to\infty} \frac{T(n,p_1)}{T(n,p_2)} = \frac{h(p_2)}{h(p_1)}, $$
and the convergence is polynomially fast.
