Can the fundamental theorem of calculus be proved without an appeal to mean value or Rolle's theorem or its immediate consequences?
As others have said, in the version of FTC in which we agree to integrate / antidifferentiate only continuous functions, the only place where a consequence of MVT occurs is when we want to assert the uniqueness of the antiderivative up to a constant, which -- as N.S. points out -- is equivalent to the Zero Velocity Theorem (ZVT):
Let $f: [a,b] \rightarrow \mathbb{R}$ be differentiable with $f'$ identically equal to zero. Then $f(x) = f(a)$ for all $x \in [a,b]$.
It is traditional to prove ZVT by noting that its contrapositive form is that a nonconstant differentiable function has a point at which the derivative is not equal to zero, and that this follows immediately from MVT. This is the way I proved ZVT in my recently completed Spivak Calculus class. Just for fun though, let me give an alternate proof using Real Induction (see e.g. page 14 of these course notes).
Note that it is no loss of generality to assume that $f(a) = 0$ and show that $f$ is identically zero on $[a,b]$. For $\epsilon > 0$, let $S(\epsilon)$ be the set of $x$ in $[a,b]$ such that for all $y \in [a,x]$, $|f(y)| \leq (x-a)\epsilon$. If we can show that $b \in S(\epsilon)$ for all $\epsilon > 0$, then we have that $|f(y)| \leq (b-a)\epsilon$ for all $y \in [a,b]$ for all $\epsilon > 0$, so $f \equiv 0$. Now:
(RI1) Since $f(a) = 0$, $a \in S(\epsilon)$.
(RI2) Suppose that for some $a \leq x < b$, $x \in S(\epsilon)$. Since $f'(x) = 0$, there exists $\delta > 0$ such that $|y-x| \leq \delta$ implies $|f(y)-f(x)| \leq
\epsilon |y-x| \leq \epsilon \delta$. Thus for all $0 < \delta' \leq \delta$ and $y \in [x,x+\delta']$,
$|f(y)| \leq |f(x)| + |f(y) - f(x)| \leq |f(x)| + \epsilon \delta' \leq (x-a)\epsilon + \delta' \epsilon = (x+\delta'-a)\epsilon$,
so $[x,x+\delta] \subset S(\epsilon)$.
(RI3): Similarly, suppose $a < x \leq b$ and $[a,x) \in S(\epsilon)$. Choose $\delta > 0$ such that $|y-x| \leq \delta$ implies $|f(y) - f(x)| \leq \epsilon |y-x| \leq \ \delta \epsilon$. For $y \in [x-\delta,x]$, we have
$|f(y)| \leq |f(y) - f(x-\delta)| + |f(x-\delta)| \leq \delta \epsilon + (x-\delta-a)\epsilon = (x-a) \epsilon$.
I can't think of a situation in which I would rather give this proof than prove the Extreme Value Theorem and use that to to prove the Mean Value Theorem, but I find arguments by real induction to have a certain charming directness to them...
I think of the FTC as a sharpening of the MVT, so it would seem artificial to me to prove the FTC without the MVT.
Consider: The MVT states that, when $f(x)$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists $c\in (a,b)$ such that the slope of the tangent line to the graph of $f(x)$ at $c$ is precisely equal to the slope of the line passing through $(a,f(a))$ and $(b,f(b))$. We don't know anything about the point c.
Wait a second! We do know something about c! We know that $f^{\prime}(c)$ is the average of the derivative between $a$ and $b$! How do we know this? Well, if the function were linear, the average would just be the slope itself; and now that it isn't linear, well, the average is still the slope of the aforementioned line. Think about it- the derivative is velocity. Sometimes the car goes faster and sometimes it goes slower, but its average velocity will be the velocity $v$ such that, if it were traveling at constant velocity $v$ all the time, it would get from $f(a)$ to $f(b)$ in time $b-a$.
Now we want a way to write down that $f^{\prime}(c)$ is this average. How do we do it? Well, the average is the total divided by the number of parts, each weighted by its length. Wait a second- the numerator of that is the Riemann integral! Taking limits, the average of $f^{\prime}$ on $(a,b)$ is $\frac{\int_{a}^b f^{\prime}(t)dt}{b-a}$. So we've rewritten the "average version" MVT as $\frac{f(b)-f(a)}{b-a}= \frac{\int_{a}^b f^{\prime}(t)dt}{b-a}$. Multiply both sizes by $b-a$ and you're finished.
To turn this into honest mathematics you use some kind of formal argument like you mentioned- but morally, there's nothing more to the FTC than the interpretation of $f^{\prime}(c)$ as not just $f^{\prime}$ evaluated at some strange point, but rather as an average.
Added: Because other answers are saying yes, I should stress that I'm saying no. There's an irreducible difficulty to the FTC, which is to translate between an epsilon-delta concept $\int_a^b f^{\prime}(t) dt$ and a number $f(b)-f(a)$. You have to cross that bridge somehow. And the only ways across are the IVT and the EVT, which in this context come in the guise of the MVT. I wish I had a logician to comment- but I think that, logically, I'm going to go out on a line here and assert that there can be no way to the FTC which doesn't pass through some form of the MVT.
Added 2: There are a number of different ways to formulate completeness of the real numbers. As other Pete L. Clark's and N.S.'s answers have shown, these, and techniques such as real induction, can be leveraged to yield genuinely different proofs of the FTC. So the answer turned out to be yes after all!
For an alternative approach to the FTC without the MVT, see this paper, which also contains other references:
Tucker, Thomas W.; Rethinking rigor in calculus: the role of the mean value theorem. Amer. Math. Monthly 104 (1997), no. 3, 231–240. http://dx.doi.org/10.2307/2974788; also available at http://www.math.cornell.edu/~maria/1220/ .
This article has brought a reaction:
Swann, Howard; Commentary on rethinking rigor in calculus: the role of the mean value theorem. Amer. Math. Monthly 104 (1997), no. 3, 241–245. http://dx.doi.org/10.2307/2974789; also available at the author's site: http://www.math.sjsu.edu/~swann/commall.pdf.
and also this
Scott E. Brodie; On "Rethinking Rigor in Calculus...," or Why We Don't Do Calculus on the Rational Numbers. The College Mathematics Journal , Vol. 30, No. 2 (Mar., 1999), pp. 135-138. http://www.jstor.org/stable/2687725.