Short answer: yes, there are numbers worse than $e$ and $\pi$. Almost all real numbers are worse. There are uncountably many reals, whereas any class of "nice transcendental numbers" you can think of will be only countable (and I propose exponential periods as a good class of numbers of the kind you describe).
The class of numbers which are roots of polynomials with rational coefficients are the algebraic numbers. The numbers which are definite integrals of algebraic functions with algebraic bounds are called the ring of periods. It is a larger class of numbers, including many familiar (suspected) transcendental numbers such as $\pi$, $\log 2$, $\zeta(3)$, and $\Gamma(p/q)^q$. See this nice primer by Kontsevich and Zagier.
Actually some common numbers like Euler's number $e$, the Euler-Mascheroni constant $\gamma$, and $1/\pi$ are still (suspected to be) missing in the ring of periods. So Kontsevich and Zagier go further and extend the ring to exponential periods, defined as integrals of products of exponentials of algebraic functions with algebraic functions. This gives a class of numbers that include "all algebraic powers of $e$, values of $\Gamma$ at rational arguments, values of Bessel functions, etc."
I assume that the exponential periods still don't include $\gamma$, because finally they claim that if you extend the class further by adding it, then you have "all classical constants in an appropriate sense" (whatever that means).
In this primer on exponential motives by Fresán and Jossen, according to Belkane and Brosnan, $\gamma$ is an exponential period, as witnessed by the integrals $\gamma=-\int_0^\infty\int_0^1e^x\frac{x-1}{(x-1)y+1}\,dy\,dx$ or $\gamma=-\int_0^\infty\int_1^x\frac{1}{y}e^{-x}\,dy\,dx$.
Anyway, the rational numbers are countable. The algebraic numbers are countable. The ring of periods is countable. The exponential periods are countable. And including $\gamma$ certainly still leaves you with a countable class of periods.
But the real numbers are uncountable, so it remains the case that most real numbers are not periods, not exponential periods, and cannot be written in any of these forms.
So to answer the question, yes, real numbers do get more exotic than numbers in the ring of periods or exponential periods like $e$ or $\pi$.
My understanding of statements of the form "$\Gamma$ takes values in the exponential periods at rational values" is that we would expect $\Gamma$ to take values at irrational arguments that are not exponential periods (although in general I expect proofs of such claims to be hard to come by).
So I would expect numbers like $\Gamma(\sqrt{2})$, $e^\pi$, $\zeta(\log 2)$ to be examples of computable numbers which are not exponential periods. But these would presumably not be considered “classical constants”.
Going further, even the computable numbers are countable, so most real numbers are not even computable, let alone in the ring of periods.
You can make (a field isomorphic to) $\mathbb C$ from $\mathbb Q[i]$ by adjoining $2^{\aleph_0}$ independent transcendentals and taking the algebraic closure. Easy-peasy.
$\mathbb R$ from $\mathbb Q$ is not so simple a matter.
If you just take $\mathbb Q(\zeta)$ you don't get any $\sqrt[3]\zeta$, even though all reals have cube roots. You will need to do something additional to make it exist. Adjoining more independent transcendentals will not do, because if they can create $\sqrt[3]\zeta$, they were not algebraically independent from $\zeta$ in the first place.
Be sure not to do anything as crude as requesting an algebraic closure while doing this, because you don't want to get any $\sqrt{-1}$.
Because positive reals can be characterized algebraically as "all the squares of nonzero reals", simply knowing the field operations of $\mathbb R$ is enough to reconstruct the ordering of $\mathbb R$. This means that if you succeed in constructing $\mathbb R$ by somehow "adding" transcendentals, the resulting structure will determine exactly how each of your transcendentals compares to each rational -- that is, you will end up having determined which real it is. And that means that you need to make a lot of decisions along the way that reach farther than "here is another transcendental, same as all the other ones".
Conversely: By Zorn's lemma, $\mathbb R$ has a maximal algebraically independent subset -- a transcendence basis. Every real number is root in a polynomial whose coefficients are finite rational polynomials of the basis elements, in an "essentially unique" way.
If you take any proper subset of the transcendence basis -- which may well be $2^{\aleph_0}$-sized -- and take all real roots of polynomials with coefficients that are rational polynomials of the selected elements, then you get a field that probably matches your intuitive idea of "adding" $2^{\aleph_0}$ transcendentals. It will not be all of $\mathbb R$, and not even isomorphic it $\mathbb R$. Its field structure alone will encode exactly which of the reals are missing.
On the other hand, you can also (in one sense) "go too far" and add too many transcendentals, getting a real closed field that properly extends $\mathbb R$. (For example, adjoin an infinitesimal element using a model-theoretic compactness argument, possibly followed by repeating the above construction to select exactly how many supernumerary independent transcendentals you want). Or even one that is neither a subfield nor an extension of $\mathbb R$.
Best Answer
Because reals are uncountable, but algebraics aren't.