First I'll give a fairly useless exact formula. The number of labelled trees with vertices of degree $d_1,\ldots,d_n$ is $$\binom{n-2}{d_1-1,\ldots,d_n-1}.$$ Therefore the number with $d_1=\cdots=d_t=k$ is $$T(n; k,t)=\frac{(n-2)!\,(n-t)^{n-2-tk+t}}{(k-1)!^t\,(n-2-tk+t)!},$$ where we agree that $T(n;k,t)=0$ if $n-2-tk+t<0$. Thus, by inclusion-exclusion, the number of trees with no vertices of degree $k$ is $$\sum_{t\ge 0} (-1)^t\binom nt T(n;k,t).$$
Next I'll indicate how the asymptotic value can be obtained. Fix any $\mu>0$. Let $\boldsymbol{D}=(D_1,\ldots,D_n)$ be a random variable whose components are iid Poisson variables with mean $\mu$. Then the conditional distribution of $\boldsymbol{D}$ subject to $\sum_{i=1}^n D_i=n-2$ is (exactly) the same as the distribution of $(d_1-1,\ldots,d_n-1)$ for a random tree. Note that this is independent of $\mu$.
We want to compute the conditional probability of $A$, which is the event that $D_i\ne k-1$ for $1\le i\le n$. By Bayes' Theorem (twice) we have $$P(A\mid {\textstyle\sum_i D_i=n-2}) = \frac{P(A)\,P(\sum_i D_i=n-2\mid A)}{P(\sum_i D_i=n-2)},$$ where the quantity on the left is the probability we need. On the right, $P(A)$ is easy: $P(A)=P(D_i\ne k-1)^n$ since the $D_i$s are independent. Also $P(\sum_i D_i=n-2)$ is easy since $\sum_i D_i$ has a Poisson distribution with mean $n\mu$. The only slightly tricky part is $P(\sum_i D_i=n-2\mid A)$.
Consider $X_1,\ldots,X_n$ to be iid random variables whose distribution is like Poisson with mean $\mu$ except that the value $k-1$ is omitted. Then $P(\sum_i D_i=n-2\mid A)$ is the probability that $\sum_i X_i = n-2$. To estimate this, adjust $\mu$ so that $\mathbb{E} X_i\approx (n-2)/n$ and apply a local central limit theorem.