The proof of the asymptotic formula for the partition function given by Hardy and Ramanujan was "the birth of the circle method", and used properties of modular forms. Erdös was trying to give a proof with elementary methods (he also gave a so-called elementary proof of the PNT with Selberg). He succeeded in 1942 to give such a proof, but only with "unknown" constant $a$, see here. Afterwards Newman gave a "simplified proof", and obtained also that $a=\frac{1}{4n\sqrt{3}}$, see here.
There are several "modern" references now, which give an elementary proof of the asymptotic formula for $p(n)$. Here are two references:
M. B. Nathanson: On Erdös's elementary method in the asymptotic theory of partitions.
Daniel M. Kane (was misspelled as "Cane"): An elementary derivation of the asymptotic of the partition function.
Instead of using modular forms etc. the elementary method of Erdös only uses elementary estimates of exponential sums and an induction from the identity
$$
np(n)=\sum_{ka\le n}ap(n-ka).
$$
It sounds like you are asking for the number $p_k(n)$ of partitions of a number $n$ into exactly $k$ parts. This is a standard type of restricted partition problem and can be solved as follows. Given such a partition, "rotate the Ferrers diagram," which shows that $p_k(n)$ also counts the number of partitions of a number $n$ whose largest part has size exactly $k$. Removing this part produces a partition of $n-k$ whose largest part has size at most $k$, and using this we can show that for fixed $k$ this sequence has a very nice generating function given by
$$P_k(x) = \sum_{n \ge 0} p_k(n) x^n = \frac{x^k}{\prod_{i=1}^k (1 - x^i)}$$
which in principle allows you to write down a closed form for $p_k(n)$ (again, for fixed $k$; it gets more complicated the larger $k$ is), although I won't do so. As a random example, setting $k = 5$ gives
$$\begin{align} \sum_{n \ge 0} p_5(n) x^n &= \frac{x^5}{(1 - x)(1 - x^2)(1 - x^3)(1 - x^4)(1 - x^5)} \\ &= x^5 + x^6 + 2x^7 + 3x^8 + 5x^9 + 7x^{10} + 10x^{11} + 13x^{12} + 18x^{13} + \dots \end{align}$$
which shows, for example, that the number of partitions $p_5(13)$ of $13$ into exactly $5$ parts is $18$. $p_5(n-5)$, which also counts the number $p_{\le 5}(n)$ of partitions of $n$ into at most $5$ parts, turns out to be A001401 on the OEIS.
Among other things, this generating function can be used to deduce the asymptotic growth rate of $p_k(n)$ (for fixed $k$, as $n \to \infty$), which turns out to be
$$p_k(n) \sim \frac{n^{k-1}}{(k-1)! k!}.$$
Edit: Okay, mea culpa - I said I'd write down the closed form you get from the generating function but for large $k$ I am really not sure how to do it even in a messy way. The basic strategy is to compute the partial fraction decomposition of the generating function $P_k(x)$ above. This gets very messy in general; here's what you get for small values of $k$.
For $k = 1$ there is a unique partition of $n$ with $1$ part, namely $n = n$. This corresponds to the generating function
$$P_1(x) = \frac{x}{1 - x} = x + x^2 + x^3 + \dots $$
which gives $\boxed{ p_1(n) = 1 }$ if $n \ge 1$ and $0$ if $n = 0$. Pretty straightforward.
For $k = 2$ the partial fraction decomposition is
$$P_2(x) = \frac{x^2}{(1 - x)(1 - x^2)} = \frac{1}{2(1 - x)^2} - \frac{3}{4(1 - x)} + \frac{1}{4(1 + x)}$$
(this can be done by hand) which gives the already-surprisingly-complicated
$$\boxed{ p_2(n) = \frac{n}{2} + \frac{(-1)^n - 1}{4} }.$$
This can be simplified, and proven, with a little casework as follows. The partitions into two parts are $n = i + j$ where, if $i \ge j$, we have $1 \le j \le \lfloor \frac{n}{2} \rfloor$. So we get $\boxed{ p_2(n) = \left\lfloor \frac{n}{2} \right\rfloor }$ which is equivalent to the above after squinting a bit.
For $k = 3$ the partial fraction decomposition is (according to WolframAlpha)
$$P_3(x) = \frac{-x^2 + 20x - 7}{72(1 - x)^3} - \frac{8}{1 + x} + \frac{x + 2}{9(1 + x + x^2)}$$
which gives, using that $\frac{x + 2}{1 + x + x^2} = \frac{2 - x - x^2}{1 - x^3}$ and $-x^2 + 20x - 7 = - (1 - x)^2 - 18(1 - x) + 12$, and after some simplification,
$$\boxed{ p_3(n) = \frac{6n^2 - 7}{72} - \frac{(-1)^n}{8} + \frac{\omega_3(n)}{9} }$$
where $\omega_3(n)$ is a periodic function with period $3$ that repeats $2, -1, -1, 2, -1, -1, \dots $. Already it's not entirely obvious that this is an integer, and I can't claim not to have made any computational errors, but the result has to look something like this even if I messed something up. It only gets worse from here - $p_k(n)$ has period-$d$ components for every $1 \le d \le k$. You can check out the partial fraction decomposition for $k = 4$ for yourself here.
So there's an important sense in which it's hopeless to expect a closed form for large $k$, although as above the leading-order asymptotic in $n$ for fixed $k$ is easy to calculate and the next few terms might not be so bad either. The good news is that for many purposes it doesn't matter; the generating function is powerful enough to answer many questions you might have about partitions with restricted parts.
Best Answer
It is known that the number of partitions of $n$ with largest part $k$ is the same as the number of partitions into $k$ parts (see e.g. here). Thus we can translate your question into asking about the size of the largest part of a (uniform) random partition. The asymptotic distribution of the largest part is given by the Erdős Lehner theorem ([1]) which states that ([2], equation 2.2)
$\begin{equation} \lim_{n\rightarrow\infty}P_n(\lambda \in \mathcal{P}_n : \frac{c}{\sqrt{n}}\lambda_1 − \log \frac{\sqrt{n}}{c}\leq x) = e^{−e^{−x}} \end{equation}$, where $c=\frac{\pi}{\sqrt{6}}$.
Here $\mathcal{P}_n$ is the set of partitions of $n$, $P_n$ the uniform measure on $\mathcal{P}_n$, elements (partitions) of $\mathcal{P}_n$ are denoted by $\lambda$ ,and $\lambda_1$ denotes the largest part of $\lambda$. As an aside, this limiting distribution is known as the Gumbel distribution.
In particular, if we let $\lambda_1^{(k)}$ denote the largest part of the $k$th percentile partition (ordered in increasing order of largest part), set $e^{-e^{-x}} = k$, solve for $x$ and then for $\lambda_1$ using $\frac{c}{\sqrt{n}}\lambda_1 − \log \frac{\sqrt{n}}{c} = x$, we get $x = \log (\frac{1}{\log (1/k)})$ and $\lambda_1^{(k)} = \frac{\sqrt{n}}{c}\log (\frac{\sqrt{n}}{c\log (1/k)})$.
References:
P. Erdős, J. Lehner: The distribution of the number of summands in the partition of a positive integer. Duke Math. J. Vol. 8 (1941)
Zhonggen Su: Asymptotic analysis of random partitions