(**Update.** This is a totally new answer.)

On the one hand, the comments explain that one can have a tree $T$
with continuum many branches, but with a very slow growth rate,
adding no branching nodes at almost every level, and only very
seldom allowing a node to split, and at most one node splitting on
a level at a given time. This is easy to see, since in order to
get continuum many branches, one only needs to arrange that every
node gets a chance to split eventually, and this can be done in
such a very slow manner, simply by usually doing no splitting, but
then on a very sparse set, for the $n^{th}$ splitting, making sure
that the $n^{th}$ node on the tree is below some very high
splitting node.

But conversely, let me argue that having even a very high growth
rate does not ensure that there are uncountably many branches.

We build $T$ in stages. First, we let $T$ use the full binary
branching up to some level $n_0$. Then, at this stage, we allow
only half of the nodes to branch, the ones that started out with
$1$ in the their first bit, and force the other half, the ones
beginning with $0$, to become isolated branches in the tree starting at level $n_0$,
continued only with more $0$s after $n_0$. Meanwhile, we let the
other "live" nodes, beginning with $1$, to continue splitting above $n_0$ 
until some much larger stage $n_1$. At that level, we again kill
off half of them. Namely, any node beginning with $10$ will become
isolated at $n_1$, and the others will continue splitting up until
the very much later stage $n_2$. And so on.

The general procedure is: at level $n_k$, the nodes beginning with
$1^k0$ will become isolated at level $n_k$ (which is much larger than $k$), and the nodes beginning with $1^k$
are allowed to continue branching up to level $n_{k+1}$.

The resulting tree $T$ will have only countably many branches,
since the only branches will be the isolated branches that we
forced to occur, plus the all $11111\cdots$ branch, since whenever a sequence has its first $0$ in position $k$, it will become isolated at level $n_k$.

But now, the main point, is that by choosing the levels
$n_0<n_1<n_2<\cdots$ to be very fast growing, we can ensure an extremely high growth rate for this tree. For example, at level $n_0$, we
have the full $2^{n_0}$ many nodes; at level $n_1$, we have $2^{n_1-1}$
many nodes, and at level $n_k$, I think we have $2^{n_k-k}$ many
nodes (please correct me if my calculation of this is off).

By choosing $k\mapsto n_k$ to be very fast, we can arrange that
the growth rate of $T$ is very high.

I will give some more thought to exactly how high we can make
this growth rate. Perhaps it is something like: we can exceed any $f$ that is not
$O(2^n)$. Conversely, it seems conceivable to me that any tree whose growth rate is $O(2^n)$ must have continuum many branches.